Solution:

It is not possible. Let $L(t)$ denote the length of the boundary of the gray region after $t$ squares have been colored gray. We have $L(1) = 4$ since the perimeter of the first square colored is $4$. After $t \geq 1$ squares have been colored, if we add a square that has exactly one gray neighbor, the edge shared with that neighbor disappears from the boundary but is replaced by three new edges, so $L(t+1) = L(t) + 2$. If we add a square that has exactly three neighbors, all three neighboring edges disappear from the boundary, but there will be one new edge, so $L(t+1) = L(t) - 2$. In either case, every move after the first changes $L$ by $2 \pmod{4}$.

Since $L(1)$ is a multiple of $4$, $L(t)$ will be a multiple of $4$ whenever $t$ is odd, and $L(t)$ will be congruent to $2 \pmod{4}$ whenever $t$ is even $(t \geq 2)$. Thus if all $64$ squares could be colored in, $L(64)$ would be congruent to $2 \pmod{4}$. But $L(64)$ would have to equal the perimeter of the grid, which is $32$, a multiple of $4$. This is a contradiction.