This is an attempt to complement Nick Gill's answer. If it is correct (I am not sure), it should be merged into his answer.
Proposition: In every hole-free polygon, there exists a corner square that lies in exactly one maximal subrectangle.
Proof: Observe that as one moves clockwise around the perimeter of the polygon, one must go either left or right at various stages. One obtains a sequence: $R,R,L, R, L \cdots$ and it is clear that the number of $R$'s is 4 more than the number of $L$'s. Therefore there must be a pair of two adjacent $R$'s.
Assume w.l.o.g. that there is such a pair such that the side between the corners faces west. Consider the maximal rectangle that covers the two $R$ corners (red in the pictures below). This rectangle must run into at least one eastern side. There are two cases:
Case A: The eastern side is connected to an $R$ corner adjacent to the $RR$ pair at the north or south:
In this case, the middle $R$ corner is contained in exactly one maximal rectangle (the red one) and we are done.
Case B: The eastern side is between two adjacent $L$ corners:
By the construction, it is clear that this $LL$ pair cannot be used in the same way with any other $RR$ pair.
Now we know that this cannot happen to all consecutive $R$-corners, because the number of $R$-corners exceeds the number of $L$-corners by $4$.
EDIT: I am not sure about the latter counting argument. What if there is a sequence LLL, such that the first LL pair matches an RR pair, and the second LL pair matches a disjoint RR pair?
