Let $B$ be a complete Boolean algebra such that if $p_{n}$ is a finite partition of $B$ for all $n$, then there are $b_{n}\in p_{n}$ for all $n$ where $\bigwedge_{n\in\omega}b_{n}\neq 0$. Let $X=S(B)$ be it's Stone space. Then let
$f:X\rightarrow\{0,1\}^{\alpha}$ be a continuous function. Then let $P_{n}$ be the partition of $\{0,1\}^{\alpha}$ where we set $(x_{\alpha})_{\alpha}=(y_{\alpha})_{\alpha}(P_{n})$ if and only if $x_{i}=y_{i}$ for each $i<n$. Then $P_{n}$ is a partition of $\{0,1\}^{n}$ into clopen sets, so
$Q_{n}=\{f^{-1}[U]\mid U\in P_{n}\}^{+}$ is a partition of $X$ into clopen sets. Therefore, by our hypothesis, there exists some non-empty open subset $U\subseteq X$ and open sets $U_{n}\in Q_{n}$ such that $U\subseteq\bigcap_{n}U_{n}$. Therefore, if we set $O_{n}\in Q_{n},U_{n}=f^{-1}[O_{n}]$, then $U\subseteq\bigcap_{n}f^{-1}[O_{n}]=f^{-1}[\bigcap_{n}O_{n}]$. Therefore, $f[U]\subseteq\bigcap_{n}O_{n}$, but clearly $\bigcap_{n}O_{n}$ has empty interior. Thus, $f[U]$ also has empty interior.
In other words, there is no quasi-open $f:X\rightarrow\{0,1\}^{\alpha}$.

Let $B$ be a complete Boolean algebra such that if $p_{n}$ is a finite partition of $B$ for all $n$, then there are $b_{n}\in p_{n}$ for all $n$ where $\bigwedge_{n\in\omega}b_{n}\neq 0$. Let $X=S(B)$ be it's Stone space. Then let
$f:X\rightarrow\{0,1\}^{\alpha}$ be a continuous function. Then let $P_{n}$ be the partition of $\{0,1\}^{\alpha}$ where we set $(x_{\alpha})_{\alpha}=(y_{\alpha})_{\alpha}(P_{n})$ if and only if $x_{i}=y_{i}$ for each $i<n$. Then $P_{n}$ is a partition of $\{0,1\}^{n}$ into clopen sets, so
$Q_{n}=\{f^{-1}[U]\mid U\in P_{n}\}^{+}$ is a partition of $X$ into clopen sets. Therefore, by our hypothesis, there exists some non-empty open subset $U\subseteq X$ and open sets $U_{n}\in Q_{n}$ such that $U\subseteq\bigcap_{n}U_{n}$. Therefore, if we set $O_{n}\in Q_{n},U_{n}=f^{-1}[O_{n}]$, then $U\subseteq\bigcap_{n}f^{-1}[O_{n}]=f^{-1}[\bigcap_{n}O_{n}]$. Therefore, $f[U]\subseteq\bigcap_{n}O_{n}$, but clearly $\bigcap_{n}O_{n}$ has empty interior. Thus, $f[U]$ also has empty interior.
In other words, there is no quasi-open $f:X\rightarrow\{0,1\}^{\alpha}$.

For problems like this, it is helpful to use Stone duality to turn properties of compact totally disconnected spaces into properties of Boolean algebras.

Suppose that $X$ be a compact totally disconnected space. Then a mapping $f:X\rightarrow\{0,1\}^{\omega}$ is dual to a mapping
$\phi:F_{\omega}\rightarrow B$ where $F_{\omega}$ is the free Boolean algebra on countably many generators $(x_{n})_{n}$. By freeness, the Boolean algebra homomorphisms $\phi:F_{\omega}\rightarrow B$ are in a one to one correspondence with the functions $s:\{x_{n}|n\in\omega\}\rightarrow B$. In other words, the Boolean algebra homomorphisms $\phi:F_{\omega}\rightarrow B$ are in a one-to-one correspondence with the sequences $(b=a_{n})_{n}$ of elements in $B$. Therefore, let
$b_{n}=\phi(x_{n})$ for all $n$.

Now, $f$ is quasi-open if whenever $U$ is clopen in $X$, we have $\bigcap_{n\leq N}\pi_{n}^{-1}[\{c_{n}\}]\subseteq f[U]$. Said, differently, $f$ is quasi-open iff whenever $U$ is a clopen subset of $X$, there is some sequence $(c_{n})_{n\leq N}$ of bits where if
$(d_{n})_{n\in\omega}$ is an extension of the sequence $(c_{n})_{n\leq N}$, then $(d_{n})_{n\in\omega}=f(x)$ for some $x\in X$.

Now, by Stone duality, $f$ is quasi-open if and only if for all $b\in B^{+}$, there is a sequence $(c_{n})_{n\leq N}$ where $c_{n}\in\{b_{n},b_{n}'\}$ for all $n$ such that if $(d_{n})_{n\in\omega}$ extends $(c_{n})_{n\leq N}$ and $d_{n}\in\{b_{n},b_{n}'\}$ for all $n$, then there is an ultrafilter $U\subseteq B$ such that $b\in U$ and $d_{n}\in U$ for all $n$.

We can remove the ultrafilters from the characterization of quasi-openness.

The mapping $f:X\rightarrow\{0,1\}^{\omega}$ is quasi-open if and only if
for all $b\in B^{+}$, there exists a sequence $(c_{n})_{n\leq N}$ such that $c_{n}\in\{b_{n},b_{n}'\}$ for all $n\leq N$ and where if
$(d_{n})_{n\in\omega}$ is a sequence that extends $(c_{n})_{n\leq N}$ with
$d_{n}\in\{b_{n},b_{n}'\}$ for all $n\in\omega$, then
$b\wedge\bigwedge_{n\leq M}d_{n}\neq 0$ for all $n$.

Therefore, if $X$ is a compact zero-dimensional space with $X=S(B)$ for some Boolean algebra $B$, then there exists a quasi-open map $f:X\rightarrow\{0,1\}^{\omega}$ if and only if there exists a sequence $(b_{n})_{n\in\omega}$ of elements in $B$ where for all $b\in B^{+}$, there exists an $N$ and a sequence $(c_{n})_{n\leq N}$ with $c_{n}\in\{b_{n},b_{n}^{+}\}$ for all $n\leq N$ such that whenever $(d_{n})_{n\in\omega}$ is a sequence with $c_{n}\in\{b_{n},b_{n}^{+}\}$ for all $n\in\omega$ that extends $(c_{n})_{n\leq N}$, then for all $M$, we have $b\wedge\bigwedge_{n\leq M}d_{n}\neq 0$.

Finally, there are examples of compact extremally disconnected spaces $X$ and quasi-open maps $f:X\rightarrow\{0,1\}^{\omega}$. Let $B=Ro(\{0,1\}^{\omega})$, and let $X=S(B)$. Let $C(\{0,1\}^{\omega})$ be the algebra of clopen subsets of $\{0,1\}^{\omega}$. Then clearly $F_{\omega}=C(\{0,1\}^{\omega})\subseteq\text{Ro}(\{0,1\}^{\omega})$, so let $(b_{n})_{n}$ be the canonical generators of $F_{\omega}$. Then whenever $b\in B^{+}$, there is some $N$ and $c_{n}\in\{b_{n},b_{n}'\}$ for $n\leq N$ where $0<\bigwedge_{n\leq N}c_{n}\leq b$. Therefore, there is a quasi-open mapping $f:X\rightarrow\{0,1\}^{\omega}$.

Now, recall that Sikorski's extension theorem states that if $A,B,C$ are Boolean algebras where $A\subseteq B$ and $C$ is complete, then every Boolean algebra homomorphism $f:A\rightarrow C$ can be extended to a Boolean algebra homomorphism $g:B\rightarrow C$. Therefore, by using Sikorski's extension theorem, we can obtain another characterization of the compact extremally disconnected spaces that map quasi-openly onto $\{0,1\}^{\omega}$.

Observe that if $B$ is a Boolean algebra, and $\phi:F_{\omega}\rightarrow B$, then the dual map $S(\phi):S(B)\rightarrow S(F_{\omega})$ is quasi-open if and only if for all $b\in B^{+}$, there exists a $c\in F_{\omega}^{+}$ where for all $d\in F_{\omega}^{+}$ with $d\leq c$ we have $\phi(d)\wedge b\neq 0$.

Therefore, if $B$ is a complete Boolean algebra, and $G_{\omega}$ is the completion of $F_{\omega}$, then by the Sikorski extension theorem, there exists some quasi-open continous map $f:S(B)\rightarrow\{0,1\}^{\alpha}$ if and only if there exists a Boolean algebra homomorphism
$\phi:G_{\omega}\rightarrow B$ such that for all $b\in B^{+}$, there exists a $c\in F_{\omega}^{+}$ where for all $d\in G_{\omega}^{+}$ with $d\leq c$, we have $\phi(d)\wedge b\neq 0$.