Ok.  Let's say $r \leq c$ and say $N = 2^r$.

Then the number of matrices with each column distinct is exactly $N(N-1)\cdots (N-c+1)$.  This is an upper bound on the first question you asked.

Approximation time:  suppose $r^2 \ll 2^c$ then almost all (asymptotically all) matrices will have distinct rows (birthday paradox) anyway.  Since $r \geq c$, this condition will happen provided $r \gg 1$.  Therefore, having distinct rows basically always happens, so the upper bound given above is essentially the truth [this can be made more rigorous and exact as desired].

If you want to look at the equivalence class thing, then just use the fact that most matrices will have only one automorphism anyway.  So just divide the answer by $r! c!$.

Punchline: the thing is asymptotically easy, and the answer is more or less exactly what you'd think.