Let $a:=\alpha$. Note that \begin{equation} Ee^{zX_n}=\frac{\sinh(z/n^a)}{z/n^a} \end{equation} for real $z>0$. Using the inequality $\dfrac{\sinh u}u<e^{u^2/6}$ for real $u\ne0$ and the independence of the $X_n$'s, we get \begin{equation} Ee^{zR(t)}<e^{z^2 B_{a,t}/6}, \end{equation} where \begin{equation} B_{a,t}:=\sum_{n<t}\frac1{n^{2a}}. \end{equation} So, for any real $x>0$, \begin{equation} P(R(t)\ge x)\le e^{-zx+z^2 B_{a,t}/6}. \end{equation} The latter bound on $P(R(t)\ge x)$ is minimized at $z=3x/B_{a,t}$. Thus, \begin{equation} P(R(t)\ge x)\le e^{-3x^2/(2B_{a,t})}. \end{equation}

Let $a:=\alpha$. Note that \begin{equation} Ee^{zX_n}=\frac{\sinh(z/n^a)}{z/n^a} \end{equation} for real $z>0$. Using the inequality $\dfrac{\sinh u}u<e^{u^2/6}$ for real $u\ne0$ (see e.g. this MathSE answer) and the independence of the $X_n$'s, we get \begin{equation} Ee^{zR(t)}\le e^{z^2 B_{a,t}/6}, \end{equation} where \begin{equation} B_{a,t}:=\sum_{n<t}\frac1{n^{2a}}. \end{equation} So, for any real $x>0$, \begin{equation} P(R(t)\ge x)\le e^{-zx+z^2 B_{a,t}/6}. \end{equation} The latter bound on $P(R(t)\ge x)$ is minimized at $z=3x/B_{a,t}$. Thus, \begin{equation} P(R(t)\ge x)\le e^{-3x^2/(2B_{a,t})}. \end{equation}