For a specific list of primes it is unlikely to be true unless they are concentrated in  a few congruence classes $\bmod 4d$ for some odd square-free $d.$ Even then it is unlikely.

Think first about this (easy) question. For fixed $d$ what are the primes $q$ such that $d$ is a quadratic residue $\bmod q?$ Call this set $G_d.$ We may assume that $d$ is squarefree. 

Then the members of $G_d$ are 
 - maybe $2$
 - the prime divisors of $d$ along with 
 - those primes in a union of certain congruence classes $\bmod 4d.$ Half of the classes $(r \bmod 4d)$ with $\gcd(r,4d)=1$

In some cases ($d$ even or $d$ odd with all divisors $1 \bmod 4$) it suffices to consider congruence classes $\bmod 2d$. However what is written is still correct. I  will ignore your $p$ on the assumption that the goal was to rule out $d$ being a square.

Then the specific $d$ works for a particular instance of your problem, precisely if the chosen list is one of the uncountably many infinite subsets of $G_d.$

On the other hand, suppose it is given that the members of the list (other than the divisors of $d$ in the list, if any) are chosen from some $k \ll \phi(d)$ of the congruence classes $\bmod 4d$. Then, if the $k$ are chosen at random, the chance that $d$ will work is less than $2^{-k}$. 

So starting from a list $\mathbf{q}=q_1,q_2,\cdots$ the first question is "Is there some reason to suspect that there is an $M$ so that all the members of $\mathbf{q}$ (prime to $M$) are concentrated in a few of the congruence classes $\bmod M?$" If that does not happen, then there is no hope. If it does happen for a certain $M,$ then chances still may be low.

So it very much depends on where $\mathbf{q}$ comes from. In your case of divisors of $3^{2^k}-1,$ I would guess against. There are twelve residues relatively prime to $28.$ Of these six accept $7$ as a quadratic residue and six do not. Up to $k=8,$ the twelve odd divisors of $3^{2^k}-1,$ are all $5,13,17,25 \bmod 28.$ That seems unlikely to be chance. As it turns out, $7$ is actually a quadratic non-residue relative to eleven of the twelve.

By the way, the problem of finding a $d$ which is a quadratic non-residue relative to all $q \in \mathbf{q},$ is equally difficult.