I found an elementary way to proceed through Chebyshev inequality.

Assume that $\phi\in C(\mathbb{R})$ and $|\phi(x)|\le \frac{C}{1+|x|^{1+\beta}}$ for some $\beta>0$. It is known (Carmona-Molchanov 1995) that for any $\delta>0$, $$X_t = o(|t|^\delta), t\to \infty,\tag{1}$$ a.s. It is also not hard to see that $$
\operatorname{var}\left(\frac1{2n}\int_{-n}^{n} X_t dt\right) = O\Bigl(\frac1n\Bigr), n\to\infty.
$$
Now take some $r\in (1,2)$ and $a\in (r-1,1)$ and consider $A_n = \{k n^{r-a}, k=-[n^{a}],\dots,[n^a]+1\}$. Since $r-a<1$ and thanks to $(1)$, the average does not change a lot between the points of $A_n$, so for any $\varepsilon>0$,
$$
\limsup_{n\to\infty}\mathrm{P}\left(\sup_{|x|\le n^r} \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right|>\varepsilon\right) \\
= \limsup_{n\to\infty}\mathrm{P}\left(\sup_{|x|\in A_n} \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right|>\varepsilon\right)\\
\le \limsup_{n\to\infty} \sum_{x\in A_n} \mathrm{P}\left(\sup_{|x|\in A_n} \left|\frac1{2n}\int_{x-n}^{x+n} X_t dt - \mathbb{E}[X_0]\right|>\varepsilon\right)\\
\le \limsup_{n\to\infty}\frac{cn^a}{\varepsilon^2} \operatorname{var}\left(\frac1{2n}\int_{-n}^{n} X_t dt\right)=0.
$$
(With a little bit more effort an almost sure convergence can be shown for any $r>1$, but the above is enough for me.)