We claim the answer is $m+n-1.$ This works by starting with the top-left corner, and then later placing white pieces in cells which are on either the top row or the leftmost column. Let's now show that this is optimal. We'll show this by induction on $m+n.$ If $m = 1$ or $n= 1$ the result is clearly true. Consider the set of white pieces in the end. Notice that no four of them can form a rectangle, as there is no where that we could place white pieces on all four vertices of a rectangle (consider the last white piece placed). Call a white piece a row loner if it is the only white piece in its row, and a column loner else. If there exists a row loner or a column loner, remove that row or column with only one white piece and induct down. Otherwise, suppose that no white piece is alone in its row or in its column. Consider the last white piece placed. Notice that none of the white pieces in its same row or column can be placed during the same move. However, since there is at least one other white piece in its row and in its column placed before it, that means that the cell this last white piece was placed on would have been black, contradiction. $\square$