The answer is $\inf_{p \leq \gamma_{d}(A) \leq q} \gamma(A^{\varepsilon}) = \Phi(\Phi^{-1}(p)+\varepsilon)$ where $\Phi(x) = \int_{-\infty}^{x} \frac{e^{-s^{2}/2}}{\sqrt{2\pi}}ds$.
Indeed, all you need is the following claim:
For any measurable $A \subset \mathbb{R}^{d}$ and any $\varepsilon>0$ we have $\gamma_{d}(A^{\varepsilon})\geq \Phi(\Phi^{-1}(\gamma_{d}(A))+\varepsilon)$. The equality is attained if $A$ is any hafspace $H$ with $\gamma_{d}(A)=\gamma_{d}(H)$.
Since both $\Phi$ and $\Phi^{-1}$ are increasing the answer to your questions follows from the claim.
The claim follows from the Ehrhard inequality: for any measurable sets $U, V \subset \mathbb{R}^{d}$ and any $\alpha, \beta \geq 0$ such that $\alpha+\beta \geq 0$, $|\alpha-\beta|\leq 1$, $\alpha U+\beta V$ is measurable (here "+" is Minkowski sum) we have
$$
\gamma_{d}(\alpha U +\beta V) \geq \Phi(\alpha \Phi^{-1}(\gamma_{d}(U))+\beta\Phi^{-1}(\gamma_{d}(V))).
$$
So, let $B$ be the unit ball, and $\lambda \in (0,1)$. Then by Ehrhard we have
$$
\gamma_{d}(A^{\varepsilon}) = \gamma_{d}(\lambda [\lambda^{-1}A]+(1-\lambda)[(1-\lambda)^{-1}\varepsilon B]) \geq \Phi(\lambda \Phi^{-1}(\gamma_{d}(\lambda^{-1}A))+(1-\lambda)\Phi^{-1}(\gamma_{d}((1-\lambda)^{-1} \varepsilon B)))
$$
Next, let $\lambda \to 1$, $\lambda<1$. Then $\lambda \Phi^{-1}(\gamma_{d}(\lambda^{-1}A)) \to \Phi^{-1}(\gamma_{d}(A))$, and also $(1-\lambda)\Phi^{-1}(\gamma_{d}((1-\lambda)^{-1} \varepsilon B)) \to \varepsilon$. The first limit is simple. To verify the second limit we need to show that $\lim_{r \to \infty}\frac{1}{r} \Phi^{-1}(\gamma_{d}(r B))=1$. Perhaps there are many ways to show this. The most straightforward one will be to honestly compute all asymptotic:
$$
\gamma_{d}(r B) = 1- \frac{\sigma_{d}}{(2\pi)^{d/2}}\int_{r}^{\infty}e^{-r^{2}/2}r^{d-1}dr \stackrel{r \to \infty}{=}1-\frac{\sigma_{d}}{(2\pi)^{d/2}} e^{-r^{2}/2} r^{d-2}+o(e^{-r^{2}/2} r^{d-2}),
$$
where $\sigma_{d}$ is the surface area measure of the unit sphere. Thus
$$
\Phi^{-1}(s) \stackrel{s \to 1^{-}}{=} (-2\ln(1-s))^{1/2}+o((-\ln(1-s))^{1/2}),
$$
therefore
$$
\lim_{r \to \infty} \frac{1}{r} \Phi^{-1}(\gamma_{d}(r B)) = \lim_{r \to \infty} \frac{1}{r} [-2\ln(r^{d-2}e^{-r^{2}/2})]^{1/2}=1.
$$