This is a nice question from a while ago, and this is only a very partial answer, hoping it will motivate more complete ones.

I'll address the topological equivalent reformulation. It has a positive answer in a few cases described below. First, say that $X$ is good if the question has a positive answer for this given $X$, for every $Y$.

1) If $X$ has a clopen subset homeomorphic to $X\sqcup X$ then $X$ is good. (In particular the answer to the question is positive when $Y=X$). Indeed, in this case, define by induction pairwise disjoint clopen subsets $Y_1,\dots,Y_{n-1},Y'_{n}$ of $Y$ each homeomorphic to $X$, and then choose two disjoint clopen subsets $Y_n,Y'_{n+1}$ of $Y'_n$, each homeomorphic to $X$.

Similarly, the answer to the question is positive if $Y$ has a clopen subset homeomorphic to $Y\sqcup Y$ (or to $Y\sqcup X$).

2) If $X$ embeds as a clopen subset in a finite disjoint union $nX'$ and vice versa, then $X$ is good iff $X'$ is good.

3) If $X$ is countable then $X$ is good. Indeed, the case $X$ empty being trivial, suppose $X$ nonempty and let $\alpha=\alpha_X$ be the largest Cantor-Bendixson rank of a point in $X$; it is achieved by $n$ points, where $n=n_X$ is a positive integer, the pair $(\alpha,n)$ characterizing $X=X(\alpha,n)$ up to homeomorphism. Note that $X$ is then homeomorphic to the disjoint union of $n$ copies of $X(\alpha,1)$, so, in view of (2), let us focus on the case $n_X=1$.

So the condition that $Y$ has $k$ disjoint clopen copies of $X$ means that $Y$ has $\ge k$ points with a countable neighborhood and of Cantor-Bendixson rank $\alpha$. Hence the condition that it has this for arbitrary large $k$ means that $Y$ has infinitely many points with a countable neighborhood and of Cantor-Bendixson rank $\alpha$. Let $(x_n)$ be an injective sequence of such points. By induction let $Y_n$ be a countable clopen neighborhood of $x_n$ disjoint from $\bigcup_{i<n}V_i$, and with the additional condition that every point in $Y_n\smallsetminus\{x_n\}$ has Cantor-Bendixson rank $<\alpha$ (in particular, $x_i\notin Y_n$ for every $i>n$). Then $Y_n$ is homeomorphic to $X$ for each $n$, and these are pairwise disjoint clopen subsets.

4) Call $X$-point a point in a topological space a point with a basis of neighborhoods homeomorphic to $X$. If $X$ possesses an $X$-point that is isolated among $X$-points, then then $X$ is good (call this isolated $X$-point). This is actually a generalization of (3) since countable spaces $X$ with $n_X=1$ have this property (the point of maximal Cantor-Bendixson rank being an isolated $X$-point). The argument follows that of (3). An example not covered by the previous items is the case of the space obtained as union of a Cantor space and a discrete countable set accumulating at a single point of this Cantor space.

5) I don't know if $X,X'$ good imply $X\sqcup X'$ good; however this is clear if $X,X'$ have no nonempty homeomorphic subsets. This applies to $X$ Cantor and $X'$ countable.

Actually the few metrizable Stone spaces I can think of, so far, seem to satisfy close variants of the latter arguments. For the classification of metrizable Stone spaces, see notably

* Reichbach, M.
The power of topological types of some classes of 0-dimensional sets.
Proc. Amer. Math. Soc. 13 1962 17-23* (Open link).