**Too long for a comment:**

So you have $$M=\sum_{k=1}^{n-1}\binom{2^n}{k}\leq \binom{2^n}{n}$$ (for $n$ large enough) such possible (nonempty) sets $S$ and there are 
$$
F=
\frac{ (q^m-1)(q^m-q) \cdots (q^m-q^{k-1})}{(q^n-1)(q^n-q) \cdots (q^n-q^{k-1}) }
$$
distinct $n-$dimensional subspaces in the target space. Thus by a greedy method it looks like if $m$ is large enough for $M\geq F,$ to hold such a mapping can be found.