GH from MO gave in his answer a bound due to himself and Valentin Blomer of $O(\sigma(n))$. I thought it would be interesting to compute the $O$. Here I'm looking for an effective bound of the form $\leq C \sigma(n) + o(\sigma(n))$ where the $C$ is explicit but the little $o$ need not be, so we only need be concerned with the Eisenstein term.
The formula in the cited paper is $$\frac{\pi^2 n}{d} \prod_p \chi(p) \leq \frac{\pi^2 n}{\sqrt{d}} \cdot 4 d_0^{1/6} 2^{ \omega(d_1)} \prod_{p \geq 2} (1+p^{-2}) \frac{\sigma(n)}{n} $$
where $d$ is the discriminant, $d_0d_1=d$ and $d_1$ is prime to $d$, so $2^{\omega(d_1)}\leq (2/\sqrt{3}) d_1^{1/2}$
$$\leq \frac{8}{\sqrt{3}} \pi^2 \prod_{p \geq 2} (1+p^{-2}) \sigma(n)$$
Now $$\prod_{p \geq 2} (1+p^{-2}) = \frac{ \zeta(2)}{\zeta(4)} = \frac{15}{\pi^2} $$
so we get the explicit bound for the main term of $40 \sqrt{3} \sigma(n)$. However this is likely not sharp. Because the bound goes to $0$ with the discriminant, one could in principle write down a finite list of forms capable of ever getting above $30 \sigma(n)$, say, and try to explicitly calculate the greatest possible ratio of the main term to $\sigma(n)$ for each one.
Also this bound is not so bad. It's only this weird factor of $2/\sqrt{3}$, times another factor of $2$, above the conjectured best possible.