Since $g=(d_1,d_2)$ we have $g=rd_1+sd_2$ for some integers $r$, $s$. The general expression for $g$ is $g=(r+td_2/g)d_1+(s-td_1/g)d_2$, where $t$ is arbitrary.

From $m\equiv n\pmod g$ we get $$m=n+gk=n+k((r+td_2/g)d_1+(s-td_1/g)d_2)$$ for some integer $k$. We want to know if we can choose $t$ so that $$(n+k(r+td_2/g)d_1,q)=1$$ To put it another way, this is $(u+tv,q)=1$, where $u=n+krd_1$ and $v=kd_1d_2/g$.

This will be impossible if there is a prime $p$ which divides $u$, $v$, and $q$, that is, a prime dividing $n+krd_1$, $kd_1d_2/g$, and $q$. Such a prime can't divide $k$ or $d_1$, since $(n,q)=1$, but I don't see how to rule out the possibility that it divides $n+krd_1$, $d_2/g$, and $q$.

If we can rule out that possibility, then we can proceed: let $t$ be the product of all the primes that divide $q$ but not $u$. Then any prime dividing $q$ divides either $u$ or $tv$, but not both, hence, not $u+tv$, and we're done.