Very rarely. If $GL_2(A)$ is a Galois group then $GL_1(A)$ is a Galois group.

So in particular $Hom ( GL_1(A), \mathbb Z/p)$ must be finite, because there are finitely many $\mathbb Z/p$-extensions ramified outside a given finite set of primes.

If the residue field of $A$ has characteristic $p$, then $(A/p)^\times$ had better also have finite maps to $\mathbb Z/p$, which I think implies that it has Krull dimension $0$. Then it will be finite, and $A$ will be a finite extension of $\mathbb Z_p$ by Nakayama's lemma.

If $A$ is the ring of integers of a degree $n$ extension of $\mathbb Z_p$, then $GL_1(A)$ maps to $\mathbb Z_p^n$, so $\mathbb Z_p^n$ is also a Galois group over $\mathbb Q$. But $\mathbb Q$ has a unique $\mathbb Z_p$-extension, the cyclotomic extension.

So I think just $\mathbb Z_p$ works here.

Claim: Given a representation from the Galois group to $GL_2(\mathbb F_q)$ (maybe $p>2$ to be safe) whose image contains $SL_2(\mathbb F_q)$, if $R$ is any quotient of the deformation ring of that representation (actually, the ring parameterizign deformations with fixed determinant character), then the image of the induced map to $GL_2(R)$ contains $SL_2(R)$.

Proof: By a limit we may reduce to finite length rings. Then by induction we may reduce to an extension by $\mathbb F_q$.

So let $R$ be a quotient of the deformation ring with maximal ideal $m$ and an ideal $I$ that is isomorphic as an $R$-module to $R/m$. Assume the representation to $GL_2(R/I)$ contains $SL_2(R/I)$. We must show that the representation to $GL_2(R)$ contains $SL_2(R)$.

It's sufficient to show that it contains the kernel of $SL_2(R)$ to $SL_2(R/I)$ or by conjugation sufficient to show that it contains any nontrivial element of that kernel.

If $I$ is contained in $m^2$, we can take $a,b \in m$ with $ab$ generating $I$ and take the commutator of the two matrices:

$$ \left[ \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1+b \end{pmatrix} \right] =\begin{pmatrix} 1 & ab \\ 0 & 1 \end{pmatrix} $$

By surjectivity in $R/I$, we have two matrices congruent mod $I$ to these, and their commutator will be the same. So that handles that case.

If $I$ is contained in $(p)$, you can do the same thing with

$$\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}^p = \begin{pmatrix} 1 & pa \\ 0 & 1 \end{pmatrix}$$

Otherwise $I$ corresponds to an element of $m/(p,m^2)$, which comes from a nontrivial extension class of the representation with itself. It's sufficient to show that for such a nontrivial extension there are some nontrivial elements that fix the sum and quotient. But otherwise you would get a nontrivial extension of the standard representation of $SL_2(\mathbb F_q)$ as a representation of $SL_2(\mathbb F_q)$, which there probably isn't by modular representation theory.

OK so this proof is not completely rigorous. But I think it can be made so with a bit more care.

Anyway, this would show that the question is basically equivalent to Passerby's version about quotients of deformation rings.

Very rarely. If $GL_2(A)$ is a Galois group then $GL_1(A)$ is a Galois group.

So in particular $Hom ( GL_1(A), \mathbb Z/p)$ must be finite, because there are finitely many $\mathbb Z/p$-extensions ramified outside a given finite set of primes.

If the residue field of $A$ has characteristic $p$, then $(A/p)^\times$ had better also have finite maps to $\mathbb Z/p$, which I think implies that it has Krull dimension $0$. Then it will be finite, and $A$ will be a finite extension of $\mathbb Z_p$ by Nakayama's lemma.

If $A$ is the ring of integers of a degree $n$ extension of $\mathbb Z_p$, then $GL_1(A)$ maps to $\mathbb Z_p^n$, so $\mathbb Z_p^n$ is also a Galois group over $\mathbb Q$. But $\mathbb Q$ has a unique $\mathbb Z_p$-extension, the cyclotomic extension.

So I think just $\mathbb Z_p$ works here.