Regarding the weak version, I can prove that there exists an $f$ such that for every two colouring of the $f(n) \times f(n)$ grid, every permutation of $[n]$ is contained in either $B$ or $W$. The proof leaves a lot of room for optimization, so perhaps one can get down to $f=2n$.

**Proof.**
Evidently, $f(1)=1$ works.  We then define $f$ recursively via $f(n+1)=(n+2)(f(n)+1)+1$. 
Consider a 2-colouring of the $f(n+1) \times f(n+1)$ grid $G$ and suppose that some permutation $\sigma$ of $[n+1]$ does not appear in either $B$ or $W$.  Let $\sigma'$ be the permutation of $[n]$ obtained by removing $n+1$ and $\sigma(n+1)$ from $\sigma$ and then renaming $[n+1] \setminus \sigma(n+1)$  according to their relative order.  

Let $G'$ be the subgrid of $G$ containing the entry $(1,1)$ and with horizontal and vertical entries spaced $n+1$ entries apart.  By choice of $f$ we have that $G'$ is a $(f(n)+2) \times (f(n)+2)$ grid.  Let $G''$ be the subgrid of $G'$ obtained by removing the boundary entries.  Thus, $G''$ is a $f(n) \times f(n)$ grid.  By induction, $G''$ contains a black copy of $\sigma'$ or a white copy of $\sigma'$.  Assume it is black.  Let $G_1, \dots, G_{n+1}$ be the $(n+1) \times (n+1)$ blocks of $G$ *between* the last two columns of $G'$.  Note that every entry of $G_{\sigma(n+1)}$ must be white, otherwise, $G$ contains a black $\sigma$.  But now $G_{\sigma(n+1)}$ contains a white copy of every permutation of $[n+1]$.