$\newcommand\ep\varepsilon\newcommand\ze\zeta\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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}

`\vert\vert`

should usually be`\Vert`

. Compare, for example, $\lvert\lvert f\rvert\rvert$`\lvert\lvert f\rvert\rvert`

to $\lVert f\rVert$`\lVert f\rVert`

. I have edited accordingly. $\endgroup$