$\newcommand\R{\mathbb R}\newcommand\ep\epsilon\newcommand{\de}{\delta} $For $(s,t)\in\R^2$, let
\begin{equation}
u(s,t):=\sum c_k g\Big(\frac{R-r_k}{h_k}\Big),
\end{equation}
where $g(z):=\max(0,1-|z|)$ for real $z$,
$R:=\sqrt{s^2+t^2}$,
\begin{equation}
c_k:=h_k^{1-\de/2},\quad h_k:=k^{-3/(1-\de)}, \quad r_k:=k^2/\ln^2k,
\end{equation}
and $\sum:=\sum_{k\ge k_0}$, where in turn $k_0$ is an integer large enough
so that $k\ge2$ and for all $k\ge k_0$ we have $0<r_k-h_k<r_k+h_k<r_{k+1}-h_{k+1}-2$.

Then
\begin{equation}
|\nabla u(s,t)|^2=\sum \frac{c_k^2}{h_k^2}\, 1(|R-r_k|<h_k)
=\sum h_k^{-\de}\, 1(|R-r_k|<h_k)
\end{equation}
almost everywhere (a.e.).
So, for each $x=(s,t)\in\R^2$ there is some integer $k\ge k_0$ such that for all $r\in(0,1]$
\begin{equation}
\frac1{|B_r|}\int_{B_r(x)} |\nabla u|^2
\ll \frac1{r^2} h_k^{-\de}\,\min(h_k,r)r\le \frac1{r^\de};
\end{equation}
here and in what follows, $A\ll B$ means $A\le CB$ for some universal real constant $C>0$. So, the condition displayed in the OP holds (up to a universal positive real constant factor, which can obviously be removed by rescaling $u$).
Also,
\begin{equation}
\int_{\R^2} |\nabla u|^2\ll \sum \frac{c_k^2}{h_k^2}\,r_k h_k
=\sum h_k^{-\de}\,r_k h_k=\sum\frac1{k\ln^2k}<\infty
\end{equation}
and
\begin{equation}
\int_{\R^2} |u|^2\ll \sum c_k^2\,r_k h_k\le\sum \frac{c_k^2}{h_k^2}\,r_k h_k <\infty,
\end{equation}
so that $u\in W^{1,2}(\R^2)$.

Thus, all the conditions on $u$ hold. However, for any real $\ep>0$ there some real $\eta>0$ such that
\begin{equation}
\int_{\R^2} |\nabla u|^{2+\ep}\asymp \sum \frac{c_k^{2+\ep}}{h_k^{2+\ep}}\,r_k h_k
=\sum h_k^{-(1+\ep/2)\de}\,r_k h_k
=\sum r_k h_k^{(1-\de)(1-\eta)}=
\sum\frac1{k^{1-3\eta}\ln^2k}=\infty,
\end{equation}
so that $\nabla u\notin L^{2+\ep}$. (This answers the original version of the question, before the replacement of $L^{2+\ep}$ by $L^{2+\ep}_{loc}$.)

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