The answer is $43$.

First, we prove that at least $43$ sticks must be removed. Suppose the figure does not contain any rectangles, then each bounded connected area must consist of at least three unit squares, i.e., the area is at least $3$. Therefore, there can be at most $\lfloor \frac{64}{3} \rfloor = 21$ bounded connected areas. Removing one stick can at most reduce the number of bounded connected areas by $1$ (either by merging two bounded areas into one, or by merging a bounded area with an unbounded area). Initially, there are $64$ bounded connected areas, so at least $64 - 21 = 43$ sticks must be removed.

The figure below shows an example of removing $43$ sticks, where each bounded connected area has an area of $3$, and the figure does not contain any rectangles.