$\newcommand\ep\varepsilon\newcommand\ze\zeta\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer is yes.
Indeed, take any real $\be>0$. Let
\begin{equation*}
\al:=\be/2,\quad\ep:=\be^2/48,\quad\ze:=\eta:=\be/4.
\end{equation*}
Write $B_x(r):=[0,1]\cup(x-r,x+r)$ instead of $B_r(x)$.
Without loss of generality (wlog), $|f_n|\le M$ on $E$ for some real $M>0$ and all $n$.
By the regularity of the Lebesgue measure, there is a compact subset $K_\al$ of $E$ such that
\begin{equation*}
|E\setminus K_\al|=|[0,1]\setminus K_\al|\le\al, \tag{0}\label{0}
\end{equation*}
where $|A|$ denotes the Lebesgue measure of a subset $A$ of $\R$.
By the main condition in the OP,
\begin{equation*}
\forall x\in E\ \exists \de_{x,\ep}\in(0,\infty)\ \forall r\in[0,3\de_{x,\ep}]\ \forall n\
\end{equation*}
\begin{equation*}
\int_{B_x(r)}|f_n(y)-f_n(x)|\,dy\le\ep|B_x(r)|. \tag{1}\label{1}
\end{equation*}
Since $K_\al$ is compact, there is a finite set $G_{\al,\ep}\subset K_\al$ such that
\begin{equation*}
K_\al\subseteq\bigcup_{x\in G_{\al,\ep}}B_x(\de_{x,\ep}).
\end{equation*}
Moreover, by the Vitali covering lemma,
there is a finite set $F_{\al,\ep}\subseteq G_{\al,\ep}$ such that
the balls $B_x(\de_{x,\ep})$ for $x\in F_{\al,\ep}$ are pairwise disjoint and
\begin{equation*}
K_\al\subseteq\bigcup_{x\in F_{\al,\ep}}B_x(3\de_{x,\ep}). \tag{1.5}\label{1.5}
\end{equation*}
By \eqref{1} and Markov's inequality,
\begin{equation*}
|A_{n,r,x,\eta}|\le\frac\ep\eta\,|B_x(r)|
\end{equation*}
for all natural $n$, all $x\in F_{\al,\ep}$, and all $r\in[0,3\de_{x,\ep}]$, where
\begin{equation*}
A_{n,r,x,\eta}:=\{y\in B_x(r)\colon|f_n(y)-f_n(x)|\ge\eta\}.
\end{equation*}
So, recalling that the balls $B_x(\de_{x,\ep})$ for $x\in F_{\al,\ep}$ are pairwise disjoint, for
\begin{equation*}
A_{n,\ep,\eta}:=\bigcup_{x\in F_{\al,\ep}}A_{n,3\de_{x,\ep},x,\eta}
\end{equation*}
we have
\begin{equation*}
|A_{n,\ep,\eta}|\le\sum_{x\in F_{\al,\ep}}\frac\ep\eta\,|B_x(3\de_{x,\ep})|
\le3\frac\ep\eta\,\sum_{x\in F_{\al,\ep}}|B_x(\de_{x,\ep})|\le3\frac\ep\eta. \tag{2}\label{2}
\end{equation*}
Recalling that $|f_n|\le M$ on $E$ for all $n$ and $F_{\al,\ep}\subset E$, and passing to a subsequence if needed, wlog we have
$f_n(x)\to g(x)\ \forall x\in F_{\al,\ep}$
(as $n\to\infty$), where $g$ is some real-valued function on $F_{\al,\ep}$, so that for some natural $n_{\al,\ep,\ze}$ we have
\begin{equation*}
n\ge n_{\al,\ep,\ze}\implies\forall x\in F_{\al,\ep}\ |f_n(x)-g(x)|\le\ze.
\end{equation*}
So, if $m,n\ge n_{\al,\ep,\ze}$ and
$y\in B_x(3\de_{x,\ep})\setminus A_{m,\ep,\eta}\setminus A_{n,\ep,\eta}$ for some $x\in F_{\al,\ep}$, then
\begin{equation*}
|f_m(y)-f_n(y)|\le|f_m(y)-f_m(x)|+|f_m(x)-g(x)|+|g(x)-f_n(x)|+|f_n(x)-f_n(y)|
\le\eta+\ze+\ze+\eta,
\end{equation*}
whence, in view of \eqref{1.5},
\begin{equation*}
|f_m(y)-f_n(y)|\le2\eta+2\ze=\be
\end{equation*}
if $m,n\ge n_{\al,\ep,\ze}$ and
$y\in K_\al\setminus A_{m,\ep,\eta}\setminus A_{n,\ep,\eta}$.
So,
\begin{equation*}
|\{x\in[0,1]\colon |f_m(y)-f_n(y)|>\be\}|\le|[0,1]\setminus K_\al|
+|A_{m,\ep,\eta}|+|A_{n,\ep,\eta}|
\le\al+2\times3\frac\ep\eta=\be
\end{equation*}
if $m,n\ge N_\be:=n_{\al,\ep,\ze}=n_{\be/2,\be^2/48,\be/2}$.
So, the sequence $(f_n)$ is Cauchy convergent in measure, and hence convergent in measure. So, a subsequence of $(f_n)$ is convergent almost everywhere, as claimed.
An almost the same proof will work for the corresponding general statement for functions $f_n$ on $[0,1]^d$ for any natural $d$ and, even more generally, for any complete separable metric space $S$ with a finite doubling Borel measure $\mu$ over $S$, so that $\mu(B_x(3r))\le C\mu(B_x(r))$ for some real $C>0$, all $x\in S$, and all real $r>0$, where $B_x(r)$ is, of course, the ball in $S$ of radius $r$ centered at $x$.
Also, the main condition in the OP can be relaxed to the following:
\begin{equation}
\forall x\in E\ \forall\ep>0\ \exists\de>0\ \forall n
\end{equation}
\begin{equation}
\int_{B_x(\de)} |f_n (x)-f_n (y)|\,dy<\ep|B_x(\de)|.
\end{equation}
Best Answer
Yes I think this is true.
Let $\mathcal{G} := \{ \times_{i=1}^n[k_i,k_i+1] : k_i\in \mathbb{Z} \} $ be the grid of cubes of side length $1$ and vertices in $ \mathbb{Z}^n$. For $Q\in \mathcal{G} $ let $2Q$ be the cube with same center but double the side length. Notice that for $\varepsilon < \frac 12$ if $f$ is supported in $2Q$ then $M_\varepsilon f$ is supported in $3Q$. Let also $\phi \in C^\infty(\mathbb{R}^n), 0\leq \phi \leq 1$ supported in $ [-1,1]^n $ and $ \phi(x) = 1, \forall x \in [-\frac 12, \frac 12]^n $ and $ \phi_Q(x) = \phi(x-c_Q)$, where $c_Q$ is the center of the cube $Q$. Then for $f\in W^{1,1}(\mathbb{R^n})$ we have \begin{align*} \Vert M_\varepsilon f \Vert_{L^1(\mathbb{R}^n)} & \leq \sum_{Q\in \mathcal{G}} \Vert M_\varepsilon (f \phi_Q ) \Vert_{L^1(\mathbb{R}^n)} \\ & = \sum_{Q\in \mathcal{G}} \Vert M_\varepsilon (f \phi_Q ) \Vert_{L^1(3Q)} \\&\leq C_n \sum_{Q\in \mathcal{G}} \Vert M_\varepsilon (f \phi_Q ) \Vert_{L^{\frac{n}{n-1}}(3Q)} \\ & \leq C_n \sum_{Q\in \mathcal{G}} \Vert f \phi_Q \Vert_{L^{\frac{n}{n-1}}(\mathbb{R}^n)} \\ & \leq C_n \sum_{Q\in \mathcal{G}} \Vert f \phi_Q \Vert_{W^{1,1}(\mathbb{R}^n)} \leq C_n \Vert f \Vert_{W^{1,1}(\mathbb{R}^n).} \end{align*} The constant $C_n$ is a dimensional constant which is different in each occurance. Passing from the third to the fourth line you use the boundedness of the maximal function and the second to last inequality is Sobolev's embedding theorem.