I think the answer is YES .
Suppose that $d$ is a little square that lies in a unique maximal subrectangle $C$ in the polygon. Then any anti-rectangle contains a unique little square in $C$. We can replace this square by $d$ and still have an anti-rectangle.
So one is required to prove the following:
Proposition: Some corner square must lie in a unique maximal subrectangle.
Sketch of proposed 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.
Motivating observation: suppose that in this sequence one has three $R$'s in a row. Then the corner corresponding to the middle $R$ will lie in a unique maximal subrectangle.
The problem is that one may not have three $R$'s in a row - one does, however, have at least $2$ $R$'s in a row. Let us examine whether or not one of the two corresponding corners lies in a unique maximal subrectangle.
(Here my sketch will get icky, because I don't have Joseph O'Rourke's excellence in drawing pictures...)
I claim that the only way both of these $R$-corners can fail to lie in a maximal sub-rectangle, is if one looks at the piece of the polygon 'opposite' the corners, then one has some kind of crenellation, i.e. two knobs sticking out opposite each corner. These might not be 'smooth' - there could be many many bends in them but, still, if one gives this a little thought it becomes quite clear that this can only happen if `opposite' the edge between the two consecutive $R$-corners one has two consecutive $L$-corners.
Now we know that this cannot happen to all consecutive $R$-corners, because the number of $R$-corners exceeds the number of $R$-corners by $4$.
QED