That depends on what you mean by "codimension", but I think most interpretations give an answer of "no". For instance, let $m=n$ be $2$, let $F_0$ be $p(px_0-1)$, let $F_1$ be $(px_0 - 1)x_1$, and let $F_2$ be $(px_0-1)x_2$. Then the codimension of $\langle F_0,F_1,F_2 \rangle$ in $\mathbb{Q}[x_0,x_1,x_2]$ equals $1$; indeed, the ideal is just the principal prime ideal $\langle px_0 - 1 \rangle$. However, the codimension of $\langle F_0,F_1,F_2 \rangle$ in $(\mathbb{Z}/p\mathbb{Z})[x_0,x_1,x_2]$ equals $2$; indeed, the ideal is just the prime ideal $\langle x_1,x_2 \rangle$.

**Edit**. I should make clear that, in the above, by "codimension" of an ideal, I mean what is usually called "height", i.e., the minimal height of a prime ideal that contains the ideal (assuming the ideal is not the entire ring).