This is true, because we can approximate $f$ in $L^1$ by continuous functions. For them the statement clearly holds, and for the difference we can apply the classical Hardy-Littlewood maximal inequality. Next, I will add some details.
Define the operator $T_n$ on functions in $L^1(\mathbb R)$ by $$(T_nf)(x) := \frac{1}{2g_n (x)} \int_{x - g_n (x)}^{x + g_n (x)} f(y) \, dy.$$
By [1], $T_n$ maps $L^1$ to functions with finite weak L^1 norm, see [2] for the definition.
Given $f \in L^1(\mathbb R)$ and $\epsilon>0$, there exists a continuous function $h$ with compact support, $h \in C_c(\mathbb R)$, such that $\|f-h\|_1<\epsilon.$ Let $\mu$ be Lebesgue measure. By [1], for all $\lambda>0$, $$\mu\{x \in \mathbb R: |T_n(f-h)(x)|>\lambda/2 \} \le \frac{2C_1 \epsilon}{\lambda} \,, \tag{1}$$ where $C_1$ is an absolute constant.
Since $h$ is uniformly continuous and $g_n \to 0$ in probability, the bounded convergence theorem gives that $\|T_n h-h\|_1 \to 0$ as $n \to \infty$. Thus for all $n>n_\epsilon$, we have $\|T_n h-h\|_1<\epsilon$, so $$\mu\{x \in \mathbb R: |T_n(h)(x)|>\lambda/2 \} \le \frac{2\epsilon} {\lambda}\,. \tag{2}$$
By (1) and (2), for all $n>n_\epsilon$, $$\mu\{x \in \mathbb R: |T_n(f)(x)|>\lambda \} \le \frac{(2+2C_1)\epsilon} {\lambda}\,,\tag{2}$$ as needed.
[1] https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function
[2] https://en.wikipedia.org/wiki/Lp_space#Weak_Lp