Can a rectangle tiled by squares be written as a union of two rectangular subtilings?

Suppose I have a rectangle $R$ tiled by finitely many squares - i.e. written as a finite almost disjoint union $$R = \bigcup_{i=1}^n S_i$$ where each $S_i$ is a square and such that $S_i^\circ \cap S_j^\circ = \varnothing$ whenever $i \neq j$ (so that the squares can only overlap at their edges).

Is it always possible to write $R$ as an (almost disjoint) union $$R = R_1 \cup R_2,$$ where $R_1$ and $R_2$ are rectangles formed from the squares $S_i$? See the badly drawn image below for an example (using your imagination to turn the wonky rectangles into squares)

$\hskip1.5in$ (This question was inspired by Can a rectangle be written as a finite almost disjoint union of squares? - it follows from the discussion there that the side lengths of R are necessarily commensurable. I have no idea what the answer to this question is - it seems like something that should be false, but I can't come up with a counterexample!)