For $G$ finite the answer is yes.
It is clearly enough to know which finite groups $H$ are quotients of
$G$, as $G$ will the biggest of them.
For this we determine the number of surjective homomorphisms $G \to H$
by induction: For $H=1$ there is exactly one.
So we may assume that we know the number of surjective homomorphisms
$G \to U$ for all proper subgroups $U$ of $H$.
By assumption we are given the number $n_H$ of homomorphisms
$\varphi:G \to H$ up to conjugation by elements $h$ of $H$ where
$\varphi^h(g) = (\varphi(g))^h$.
Let $U = \varphi(G)$ be the image of $\varphi$ in $H$.
For $h\in C_H(U)$ centralizing $U\le H$ we have $\varphi^h = \varphi$,
and vice versa the equality implies that $h$ centralizes $U$.
So counting modulo conjugation is the same as weighing each homomorphism
with one over the index of the centralizer of its image.
(For $h\in N_H(U)$ normalizing $U\le H$ we know $\varphi^h(G)=\varphi(G)$.
For general $h\in H$ we get that that the image of $G$ in $H$ under
$\varphi^h$ is a subgroup of $H$ conjugated to $U$: $\varphi^h(G) = U^h$.)
Writing the number $n_H$ of homomorphisms (up to conjugacy) as
$$n_H = \sum_{\varphi:G\to H} \frac{|C_H(\mathrm{Im}(\varphi))|}{|H|}
= \sum_{U\le H}\sum_{{\varphi:G\to U} \mbox{ surjective}}
\frac{|C_H(U)|}{|H|}$$
the left side $n_H$ is known by assumption.
By induction all but one term (for $U=H$) of the outer sum on the right side can be
calculated, which enables us to determine the missing term
$x = \sum_{{\varphi:G\to H} \mbox{ surjective}} \frac{|C_H(H)|}{|H|}$. The number of surjective homomorphisms $G\to H$ is then $x$ times the index of the center of $H$.
The proof can easily be adapted for the case without conjugation mentioned in edit 2.
