The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire property whose nullsets are exactly the meagre sets and which is inner regular with respect to compact subsets. Therefore $(X, \mathcal{BP}(X), \mu)$ (where $\newcommand{\BP}{\mathcal{BP}}\BP(X)$ is the $\sigma$-algebra of sets with the Baire property) is a compact semi-finite measure space, so we can apply Fremlin's Lemma 451P. Specialized to this case, this states that if $(E_i)_{i \in I}$ is a pairwise disjoint family of sets in $\BP(X)$ such that for each $J \subseteq I$ we have $\bigcup_{i \in J}E_i \in \BP(X)$, then $\mu\left(\bigcup_{i \in I}E_i\right) = \sum_{i \in I}\mu(E_i)$. In particular, if each $E_i$ is a meagre set, (so $\mu(E_i) = 0$ for all $i \in I$), then $\bigcup_{i \in I}E_i$ is meagre.

Fremlin's proof of 451P involves an interesting argument based on ruling out the only possible counterexamples that could occur if there were an atomless or an atomic real-valued measurable cardinal involved. However, I there are other proofs, such as the following. I use the following lemma:

Lemma 1: If $X,Y$ are hyperstonean, $f : (X,\BP(X)) \rightarrow (Y,\BP(Y))$ is measurable, and $f$ reflects meagre sets (i.e. the preimage of a meagre set is meagre) then $f$ represents a complete Boolean homomorphism $\newcommand{\Cl}{\mathrm{Cl}}\Cl(Y) \rightarrow \Cl(X)$.

In the above, $\Cl(X)$ is the complete Boolean algebra of clopen sets, and we use the isomorphism between the algebra of clopen sets and sets with the Baire property modulo meagre sets to get a map $\Cl(Y) \rightarrow \Cl(X)$ from $f$. There are several ways to prove this, one of which is goes via the theory of von Neumann algebras, as follows. Integration against a complex finite measure $\nu : \BP(X) \rightarrow \mathbb{C}$ defines a normal linear functional on $L^\infty(X)$ iff $\nu$ vanishes on meagre sets. Under this isomorphism, the pushforward map $f_*$ defines a linear map $(L^\infty(X))_* \rightarrow (L^\infty(Y))_*$ which is an adjoint to the composition mapping $L^\infty(Y) \rightarrow L^\infty(X)$ defined by $f$ in the following sense: $$ \int_Y b \mathrm{d}\!f_*(\nu) = \int_X (b \circ f) \mathrm{d}\nu. $$ This implies that the composition mapping $L^\infty(Y) \rightarrow L^\infty(X)$ is weak-* continuous, and therefore defines a homomorphism of complete Boolean algebras when restricted to projections, which gives us our complete Boolean homomorphism $\Cl(Y) \rightarrow \Cl(X)$. $\Box$

To approach the question, let $X$ be a non-empty hyperstonean space and suppose for a contradiction that there exists $(E_i)_{i \in I}$ a partition of $X$ into meagre sets such that for all $J \subseteq I$, $\bigcup_{i \in J}E_i$ has the Baire property. Then the map $f : X \rightarrow I$ that maps each point $x$ to the unique $i \in I$ such that $x \in E_i$ is measurable from $(X,\BP(X))$ to $(I,\mathcal{P}(I))$. We then use the (trivially $\BP$-measurable) inclusion $i : I \rightarrow \beta(I)$, and postcompose with it, getting $f \circ i$, which is measurable $(X,\BP(X)) \rightarrow (\beta(I),\BP(\beta(I))$. The space $\beta(I)$ is hyperstonean, and the map $f \circ i$ reflects meagre sets because every meagre subset of $\beta(I)$ is contained in the set of nonprincipal ultrafilters. Therefore Lemma 1 applies and $f \circ i$ defines a complete Boolean homomorphism $\Cl(\beta(I)) \rightarrow \Cl(X)$. But for each $i \in I$, $(f \circ i)^{-1}(\{i\})$ is a meagre set, so maps to the empty set in $\Cl(X)$, but $(f \circ i)^{-1}(\beta(I)) = f^{-1}(I) = X$, a contradiction.

I think that if Lemma 1 could be proved with $X$ merely stonean, then we could use the Gleason cover of a compact Hausdorff space to show that the result holds for compact Hausdorff spaces. However, my attempts to prove Lemma 1 for stonean spaces have simply pulled me around in a circle, getting back to the same question again.

The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire property whose nullsets are exactly the meagre sets and which is inner regular with respect to compact subsets. Therefore $(X, \mathcal{BP}(X), \mu)$ (where $\newcommand{\BP}{\mathcal{BP}}\BP(X)$ is the $\sigma$-algebra of sets with the Baire property) is a compact semi-finite measure space, so we can apply Fremlin's Lemma 451Q. Specialized to this case, this states that if $(E_i)_{i \in I}$ is a pairwise disjoint family of sets in $\BP(X)$ such that for each $J \subseteq I$ we have $\bigcup_{i \in J}E_i \in \BP(X)$, then $\mu\left(\bigcup_{i \in I}E_i\right) = \sum_{i \in I}\mu(E_i)$. In particular, if each $E_i$ is a meagre set, (so $\mu(E_i) = 0$ for all $i \in I$), then $\bigcup_{i \in I}E_i$ is meagre.

[Removed wrong suggestion for a different proof.]

In the absence of the axiom of choice, it is consistent that there is a counterexample to the question for compact Hausdorff spaces. An example is given by the partition of $[0,1]$ into singletons in Shelah's model where all subsets of $\mathbb{R}$ have the Baire property.