The Boston-Markin Conjecture for Three-Manifolds

A couple of months ago I was in Oberwolfach, where Tomer Schlank told me about the following conjecture:

Conjecture. (Boston-Markin) Let $G$ be a finite group.  Then there exists a Galois extension of $\mathbb{Q}$ with Galois group $G$ ramified over exactly $d(G)$ primes, where $d(G)$ is the minimal number of generators of $G^{\text{ab}}$.  (Here we violate convention and say that $d(\{1\})=1$.)

It's not hard to see that $d(G)$ is a lower bound for the minimal number of primes over which a $G$-extension ramifies, but I don't see much evidence for the conjecture (except for some results in the case of e.g. solvable groups).  Here is a topological analogue (translating the conjecture through the usual primes-vs.-knots dictionary).

Question. Let $G$ be a finite group and $M$ a $3$-manifold.  Does there exist a link $L$ in $M$ with $d(G)$ components, and a $G$-cover of $S^3$  branched only over $L$?

The case most analogous to the original conjecture is $M=S^3$ (or in general, the case where $M$ is simply connected), where one can see that $d(G)$ is a lower bound for the number of components of a link with the desired properties.  We weren't able to make much progress with $S^3$, but Tomer, Brian Lawrence, and I were able to show that the answer is yes if $G$ is a finite simple group and $M$ is $S^2\times S^1$, or indeed any surface bundle over a circle!  (Of course such manifolds have infinite fundamental group, so the problem is strictly easier in this case.)  If I have time, I'll sketch the argument in another post.  Please let me know if you have any thoughts about this question!