If Alice selects a square $(i,j)$, Bob should choose $(i+2,j)$. In this way he will never complete a square before Alice does, and hence Bob will always win. Of course, modulo 4 is required with this strategy.
Likewise a strategy $(i,j+2)$ would have worked. Note that these strategies are always possible, because they fill the full grid with particular pairs.
In the case of a $n \times n$ grid for odd values of $n=2 m + 1$, there is a winning strategy for Alice. If she starts with $(m,m)$ as the first square and responds to any choice $(i,j)$ of Bob by playing $(2 m -i,2 m-j)$, Bob will make the first $2\times2$ square. This approach can also be generalised to rectangular grids, and will result in a winning strategy for Alice when both dimensions are odd.
Thanks to @Philip for pointing out that the pairing $(i,j)$ and $(i+2,j+2)$ would not work in the $4\times4$ grid, because the choosing of the 4 corners by Alice would create a central square made by Bob.
Thanks to @Carmeister for pointing out that a generalisation for even values $n=2m$ by pairing the square $(i,j)$ of Alice by Bob taking $(i+m,j)$, would also not be successful for the same reason. By choosing suitable squares (this time not the corners) Alice could force Bob to make a central square as well if he would stick to this particular strategy.
So the question of the existence/absence of a winning strategy for even sided square grids larger than $4\times4$ is still open.