If I am not grossly mistaken, with your notations, set $c_k := \inf_{\gamma \in \Gamma_k}\max_{u \in \gamma([0,1])} J_k(u)$ (where $\Gamma_k$ is the set of paths $\gamma : [0,1] \to H^1_0(B_k(0))$ such that $\gamma(0) = 0$ and $\gamma(1) = \varphi$ for some fixed $\varphi \in H^1_0(B_1)$ such that $J(\varphi) <0$ and $\Vert\varphi\Vert_{H^1_{0}}$ is large enough. Denote also by $c_\infty$ the analogous mountain pass value for $k=\infty$

Since upon extending a function by zero one can consider that $H^1_0(B_k(0)) \subset H^1_0(B_{k+1}(0))$, one has also $\Gamma_k \subset \Gamma_{k+1}$ and thus $0 < c_\infty \leq c_{k+1} \leq c_k$. From this one may conclude that indeed $u_k \to {\overline u} \not\equiv 0$ and that $J({\overline u})=c_{\infty}$, so that ${\overline u}$ is a non trivial solution to equation (E).

I am adding the lines below after having read the comments.

Sorry for having been so\dots elliptic in my answer.
I should have added the following details: first we choose $\varphi \geq 0$, so that the critical points $u_{k}$ are nonnegative. Then according to Gidas-Ni-Nirenberg's result $u_{k}$ has a spherical symmetry, that is $u_{k}$ is radial. Next, since
$$\Vert \nabla u_{k}\Vert^2 + \Vert u_{k}\Vert^2 = \Vert u_{k}\Vert_{p}^{p} \lesssim \Vert \nabla u_{k}\Vert^{\theta p}\Vert u_{k}\Vert^{(1-\theta)p},$$
(using Gagliardo-Nirenberg inequality, for some $0<\theta<1$) then one deduces that there is $R_{0} >0$ such that for all $k \geq1$ we have $\Vert u_{k}\Vert_{p}^{p} = \Vert \nabla u_{k}\Vert^2 + \Vert u_{k}\Vert^2 \geq R_{0}^2$. Using the fact that the imbedding $H^1_{\rm rad}({\Bbb R}^n) \subset L^{p}({\Bbb R}^n)$ is compact, one infers that $u_{k}\to {\overline u}$ strongly in $L^{p}$, and weakly in $H^1$. In particular $\Vert {\overline u}\Vert_{p}^{p}\geq R_{0}^2$, and thus ${\overline u}\not\equiv 0$. Using the equation satisfied by $u_{k}$ and the strong convergence in $L^{p}$, one checks easily that ${\overline u}$ is solution to (E), and hence
$$\Vert \nabla {\overline u}\Vert^2 + \Vert {\overline u}\Vert^2 = \Vert {\overline u}\Vert_{p}^{p},$$
yielding also that the convergence in $H^1$ is strong.