The answer is no.  Take $\alpha=\sqrt{2}$ and note that if $|\sqrt{2}-m/n|\le 1/n^2$ then we have $0<|2n^2-m^2| \le (\sqrt{2}n+m)/n \le 3$.  Now suppose we want $n\equiv 4\pmod p$ say.  Then we must have that $32-m^2 \equiv b \pmod p$ for some $|b|\le 3$.  But we can find a prime $p$ for which the numbers $29$ to $35$ are all quadratic non-residues $\pmod p$.  Thus there are no good approximations to $\sqrt{2}$ with $n\equiv 4\pmod p$ for such a prime $p$.  One can clearly vary this argument a fair bit.