This is an attempt to justify the answer $1/2$ based on the Cohen-Lenstra heuristics. There will be a lot of nonsensical steps, and I am not an expert, so this should be viewed with caution.

As is observed above, this is equivalent to determining $h(p) \mod 4$, where $h(p)$ is the class number of $\mathbb{Q}(\sqrt{-p})$. Since $p$ is odd and $3 \mod 4$, the only ramified prime in $\mathbb{Q}(\sqrt{-p})$ is the principal ideal $(\sqrt{-p})$. Thus, there is no $2$-torsion in the class group and $h(p)$ is odd.

For any odd prime $q$, let $a(q,p)$ be the power of $q$ which divides $h(p)$. We want to compute the average value of
$$\prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)}.$$

First nonsensical step: Let's pretend that the CL-heuristics work the same way for the odd part of the class group of $\mathbb{Q}(\sqrt{-p})$, that they do for the odd part of the class group of $\mathbb{Q}(\sqrt{-D})$. We just saw above that the fact that $p$ is prime constrains the $2$-part of the class group; this claim says that it does not effect the distribution of anything else.

Then we are supposed to have:
$$P(a(q,p)=0) = \prod_{i=1}^{\infty} (1-q^{-i}) = 1-1/q +O(1/q^2),$$
$$P(a(q,p)=1) = \frac{1}{q-1} \prod_{i=1}^{\infty} (1-q^{-i}) = 1/q +O(1/q^2),$$
and
$$P(a(q,p) \geq 2) = O(1/q^2).$$

If you believe all of the above, then the average value of $(-1)^{a(p,q)}$ is $ 1-2/q+O(1/q^2)$.

Second nonsensical step: Let's pretend that $a(q,p)$ and $a(q',p)$ are uncorrelated. Furthermore, let's pretend that everything converges to its average value really fast, to justify the exchange of limits I'm about to do.

Then
$$E \left( \prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)} \right) = \prod_{q \equiv 3 \mod 4} \left( 1- 2/q + O(1/q^2) \right)$$.`

The right hand side is zero, just as if $h(p)$ were equally like to be $1$ or $3 \mod 4$.