The answer is $37.$ To see that $37$ is attainable, consider the following table, where $W$ means white and $B$ means black. \[ \begin{bmatrix} W & W & B & B & W \\ B & B & W & W & B \\ W & W & B & B & B \\ B & B & W & W & B \\ W & W & B & B & W \end{bmatrix} \] Now, we will show that $37$ is optimal. For a $T-$tetromino, we will define its $\mathbf{core}$ as the cell which is in its center (the one neighboring the other three). Notice that any point is the core of at most $3$ $T-$tetrominoes, with equality iff it is surrounded by $3$ cells of its opposite color and $1$ of the same color. Furthermore, any point on the boundary is the core of at most $1$ $T-$tetromino and the corners cannot be the cores of any $T-$tetrominoes. Therefore, this gives an upper bound of $3 * 3 * 3 + 12 * 1 = 27 + 12 = 39.$ Assuming that $39$ was possible, we can WLOG assume that the center is white and then uniquely determine the other cells (up to rotation) one by one, and easily find that $39$ is not possible. Hence, we've shown that the answer is at most $38.$ Now, notice that in the case for $38,$ there is only one of the squares which is "bad," in the sense that it's the core of one too few $T-$tetrominoes. Then, caseworking on whether this bad square is on the edge or in the middle $3 \times 3$, and then splitting into subcases on where it is will finish (it's pretty straightforward, since using the non-bad squares turns out to always be enough to uniquely determine the grid). Note that an interior cell is bad iff it has $2$ neighbors of each color.