Because $R>0, u_{R}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$, $\left\|u_{R}\right\|_{\dot{H}^{1}}$ is bounded uniformly in $R$. Thus, for every $\phi \in \mathcal{D}=\left\{\right.$ smooth, compactly supported functions in $\left.\mathbb{R}^{3}\right\},$ we have \begin{equation*} \sup _{R}\left\|\phi u_{R}\right\|_{H^{1}}<\infty \end{equation*}
so $\forall R>0$, look $u_R$ as a linear operater on $\dot{H}^{1}$($u_R$ is linear when paring on test function space $\mathcal{D}$, so $u_R$ can extend to a linear operator in $\dot{H}^{1}$ by approximation), and because the dual of $\dot{H}^{1}$ is $H^{-1}$, so $u_{R}, R>0$ as a group of operater in $H^{-1}$ whcih is uniformly bounded, by Uniform boundedness principle. So to prove there is a suhbsequence of $u_{R}, R>0$ converage in $H^{-1}$, it is suffice to find a subsequence that is well-defined(as a unbounded linear operatoer in $H^{-1}$), so coverage in $L_{\text {loc}}^{p}$ for $p<6$ is suffice(I am not very sure what is your $\dot{H}^{1}$ here, if it is ${H}^{1}$, then $L_{\text {loc}}^{2}$ is suffice, the reason is if two function $f,g$ is the same as a $L_{\text {loc}}^{2}$, then they are the same measurable function, i.e. $f-g=0$ as a measurable function, so if we priori have higher regularity of $f,g$, then $f-g=0$ in the higher regularity space. ), and then it is automatally in $\dot{H}^{1}$, beacuse it is a limit of a sequence of uniformly bounded operator.
Reorganize my answer, something is wrong with the previous one,
First, an obersevation is, if there is a strong converage limit of $u_R$ in $H^1$