It turns out that this is a really simple question, but I somehow didn't think of it a while ago. We may pose the lower bound problem as an constrained optimization, whose solution is the lower bound:

$$ \min_{q'} D_{KL}(q' \parallel p)\,;\quad \text{s.t. } \int q'(x)\,\mathrm dx=1,\ \mathbb E_{q'}[x] = 0,\ \mathbb E_{q'}[x^2] = \sigma^2 \ge \gamma^2\,. $$

By letting the functional derivative of the Lagrangian w.r.t $q$ equal zero, and plugging back $q'(x) = \exp(\text{some quadratic function w.r.t } x)$ back to the constraints solving for the Lagrange multipliers, we find that the optimal $q'$ is exactly the Gaussian approximation $\bar q = \mathcal N(x \mid 0,\sigma^2)$ (see the question for its definition) of the mixture.

In conclusion, $D_{KL}(q \parallel p) \ge D_{KL}(\bar q \parallel p)$, where the r.h.s. must be greater than zero if $\gamma^2 > 1$.

EDIT: I wrote a post on this topic in detail.