Pair each empty cell with the domino that is directly to the right of this cell (unless the cell is at the right edge of the board). Suppose now that two empty cells are paired with the same domino, then this domino must be placed vertically and both cells to its left are empty. However, then there would be room for an additional domino, contradiction. Therefore, no two empty cells are paired with the same domino.

The empty cells at the right edge of the board have not yet been paired. We try to pair these cells each with a domino that has no empty cell directly to its left (and thus has not yet been paired). First, assume that it is possible to pair all empty cells at the right edge in this way with different dominoes. We have then paired all empty cells on the board with a domino, without using any domino more than once. Since each domino covers two cells of the board, for each empty cell there are two covered cells, and thus at most $\frac{1}{3}$ of the cells on the board are empty. Then we are done.

We will now show that this pairing always works. Let $k$ be the number of empty cells at the right edge and $\ell$ be the number of empty cells at the left edge. The empty cells at the left edge must not be adjacent, so there are at least $\ell-1$ dominoes at the left edge, and none of these have an empty cell to their left. If $\ell > k$, then there are enough dominoes at the left edge to pair with all empty cells at the right edge. If $\ell < k$, then we could flip the entire situation and pair all empty cells with the domino to their left, and we could also prove that at most $\frac{1}{3}$ of the cells on the board are empty. The only situation that remains is if $\ell = k$ and there are exactly $k-1$ dominoes at both edges. In this case, there is an empty cell between every two dominoes at both edges, and there is also an empty cell in the corners. But this contradicts the condition in the problem that not all corner cells are empty. This situation cannot occur.