$\newcommand\ep\varepsilon\newcommand\ze\zeta\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer is yes.
Indeed, take any real $\be>0$. Let \begin{equation*} \al:=\be/2,\quad\ep:=\be^2/144,\quad\ze:=\eta:=\be/4. \end{equation*} Write $B_x(r):=(x-r,x+r)$ 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(0,1)\ \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\le2r\ep. \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, $F_{\al,\ep}\subset[0,1]$, and $\de_{x,\ep}\in(0,1)$, 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})| =3\frac\ep\eta\,\sum_{x\in F_{\al,\ep}}|B_x(\de_{x,\ep})|\le9\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\times9\frac\ep\eta=\be
\end{equation*}
if $m,n\ge N_\be:=n_{\al,\ep,\ze}=n_{\be/2,\be^2/144,\be/2}$.
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 $x$.