I think the answer is YES .

Suppose that $d$ is a little square that lies in exactly one maximal subrectangle $C$ in the polygon. Then any anti-rectangle contains exactly one 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**: In every hole-free polygon, there exists a corner square that lies in exactly one 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. (**Edit**: as commeters have pointed out - this isn't true. But it's not used in what follows.)

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 the piece of the polygon 'opposite' the corners 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 $L$-corners by $4$.

**QED**