It is easy to prove that $\sigma(n^2)/q^k \neq q$. For suppose to the contrary that
$$q=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$
This means that
$$2n^2 = q\sigma(q^k)$$
which implies that $q \mid n^2$, contradicting $\gcd(q,n)=1$.
If the conjecture $q^k < n$ is true (see here (page 117), here, and here), then it follows that $q < n$, so that if $\sigma(n^2)/q^k < q$, then we obtain
$$\sigma(n^2) < q\cdot{q^k} < n\cdot{n} = n^2$$
which is a contradiction.
Also, if the Descartes-Frenicle-Sorli conjecture that $k=1$ is true, then it follows that $q < n$, so that if $\sigma(n^2)/q^k < q$, then we have
$$\sigma(n^2) < q\cdot{q^k} = q\cdot{q} < n\cdot{n} = n^2$$
which again is a contradiction.
Hence, conjecturally we expect
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right) > q$$
to be true.
