Solution:

2 if $mn$ is a multiple of 3, 1 otherwise

Obviously $1 \times 2$ and $2 \times 2$ can be reduced to 1. Obviously $3 \times 2$ can be reduced to 2. Note that pieces on the four $X$ squares can be reduced to a single $X$ provided that the square $Y$ is empty (call this the L move):

. . . . . . . . . . . . . . . . . . . . . . X X X X X . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . X X . . . . . . . . . . . . . . . . . . .

Thus given $m \times 2$ with $m > 3$ we can reduce it to $(m-3) \times 2$ and hence to one of $1 \times 2$, $2 \times 2$, $3 \times 2$. Note also that we are removing 3 pieces at each stage so we end up with 1 piece unless $m$ is a multiple of 3.

Given $m \times 3$ with $m > 1$ we can use the L move to reduce it to $(m-1) \times 3$. Hence by a series of L moves we get to $3 \times 1$ and hence to 2 pieces.

Now given $m \times n$ with $m \geq 4$ and $n \geq 3$, we can treat it as a $3 \times n$ rectangle adjacent to an $(m-3) \times n$ rectangle. We can now reduce the $3 \times n$ to $3 \times 3$ using L moves (with the L upright). We can then eliminate the $3 \times 3$ using L moves (with the L horizontal). Note that we have not changed $mn \bmod 3$.

This deals with all cases, except that we do not reduce $4 \times 4$ to $1 \times 4$. Instead we use L moves as follows:

X X X X X X X X X X X X X . X X . X . . . . . . . . . . . . . . . . . . . . X X X . X X X . X X X . X X . . . . . . . . . . . . . . . . . . . . X X X . X X X . X X X . X X . . . . . . . . . . . . . . . . . . . . X X X . X X X . . . . . . . . . . . . . . . . .

So we have shown that if $mn$ is not a multiple of 3 we can always reduce to a single piece. Clearly we cannot do better than that. We have also shown how to reduce to two pieces if $mn$ is a multiple of 3. It remains to show that we cannot do better. Color the board with 3 colors in the usual way:

. . . . . . . . . . . . . . . . . . . . . 1 2 3 1 2 3 . . .
. . . 2 3 1 2 3 1 . . . . . . . . . . . . . . . . . . . . 3 1 2 3 1 2 . . . . . . . . . . . . . . . . . . . .

Then any move changes the parity of the number of pieces on each color. If $mn$ is a multiple of three, then these three numbers start off equal and hence with equal parity. But a single piece has one number odd and the other two even. So we cannot get to a single piece.