The answer is negative even if $f$ is supported in $[0,1]$ and $g_n \to 0$ uniformly in $[0,1]$. The reason is that $f \in L^1$ does not suffice for the Hardy-Littlewood maximal function to be in $L^1$, so choosing $g_n$ to capture that maximal function will yield a counterexample. It is still useful to see an explicit counterexample. I will add that next.
For $x \in (0,1]$, let $$f(x)=\frac1{x \cdot \log^2(2/x)}\,,$$ with $f(x)=0$ for all $x \notin (0,1]$. Then $f$ is in $L^1[0,1]$. Define $g_n(x)= \min\{x,1/n\}$ for $x \in (0,1]$, with $g_n(x)=\frac1{n(x^2+1)}$ for $x \notin [0,1]$. Then
$$f_n (x) := \frac{1}{2g_n (x)} \int_{x - g_n (x)}^{x + g_n (x)} f(y) \, dy$$ satisfies $$f_n(x)=\frac{1}{2x\log(2/x)} \quad \text{for all} \quad x \in (0,1/n)\,,$$ so $f_n \notin L^1[0,1]$.
Remark If we assume that $f\log_+(f) \in L^1$ then the answer to the original question is positive, because in that case The Hardy-Littlewood maximal function is in $L^1$, so one can appeal to dominated convergence. See, e.g., Theorem 3.4 in this paper.
