The answer is yes, and it is based on an idea by Prokhorov (cf. e.g. Theorem 10 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). We have $EX_n^2\le Cn$ for some real $C>0$ and all natural $n$. For natural $s$, let
\begin{equation}
T_s:=\max_{2^s\le n<2^{s+1}}\frac{|X_n-X_{2^s}|}{2^s}.
\end{equation}
Then, by **[Doob's martingale inequality]**, for any real $t>0$
\begin{equation}
P(T_s\ge t)\le\frac{E(X_{2^{s+1}}-X_{2^s})^2}{(2^s t)^2}
\le\frac{EX_{2^{s+1}}^2}{(2^s t)^2}\le\frac{C2^{s+1}}{(2^s t)^2}=\frac{2C}{2^s t^2},
\end{equation}
so that $\sum_{s=1}^\infty P(T_s\ge t)<\infty$. So, by the Borel--Cantelli lemma, $T_s\to0$ almost surely (a.s.) as $s\to\infty$.
Therefore, for any natural $n$ and $r$ such that $2^r\le n<2^{r+1}$ (so that $r=r_n:=\lfloor\log_2 n\rfloor$), one has
\begin{equation}
\frac{|X_n-X_1|}n\le2 \frac{|X_n-X_1|}{2^{r+1}}
\le2\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s\to0 \tag{1}
\end{equation}
a.s., since $\sum_{s=0}^r 2^s<2^{r+1}$; cf. e.g. Lemma 9 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). Thus, $\frac{X_n}n\to0$ a.s., as desired.

Details on $(1)$: Since $T_s\to0$ a.s., without loss of generality for each real $\epsilon>0$ there is a natural-valued random variable $R_\epsilon$ such that for any natural $s$ the event $s>R_\epsilon$ implies $T_s\le\epsilon$. Therefore,

\begin{equation}
\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s
\le\frac1{2^{r+1}}\,\sum_{s=0}^{R_\epsilon} 2^s T_s+\frac1{2^{r+1}}\,\sum_{s=R_\epsilon+1}^r 2^s \epsilon
\le\frac1{2^{r+1}}\,\sum_{s=0}^{R_\epsilon} 2^s T_s+\epsilon.
\end{equation}
So, $\limsup_{r\to\infty}\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s\le\epsilon$, for any $\epsilon>0$, and hence $\lim_{r\to\infty}\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s=0$.