The answer is negative. In fact, let us show that $a_n$ as defined in the question may converge to $0$ however slowly. Indeed, let $(\epsilon_j)$ is any sequence in $(0,1]$ converging to $0$ slowly and regularly enough, in the sense that
\begin{equation}
\epsilon_{j-1}-\epsilon_j\ge2^{2-j}\tag{1}
\end{equation}
eventually in $j$ -- that is, for some natural $j_0$ and all natural $j\ge j_0$. For instance, this condition will hold if $\epsilon_j$ is (eventually in $j$) of the form $2^{2-j}$ or $j^{-a}$ or
$(\underbrace{\ln\dots\ln}_N j)^{-a}$ for some real $a$ and some natural $N$.

In what follows, $j$ is a natural number $\ge j_0$.

Let
$$k_j:=\lfloor2^j\epsilon_j\rfloor$$
and
$$r_j:=k_j/2^j.$$ Then
\begin{equation}
\epsilon_j-\tfrac1{2^j}<r_j\le\epsilon_j\tag{2}
\end{equation}
and hence, by the regularity condition $(1)$, $r_j<r_{j-1}$. Let $$\Delta_j:=(r_{j},r_{j-1}].$$
Note that the intervals $\Delta_j$ are pairwise disjoint.
Let then
$$s_{j,\ell}:=r_{j}+\ell/2^{j+1}$$
for $\ell=0,\dots,\ell_j:=(r_{j-1}-r_{j})2^{j+1}$, so that $\ell_j$ is even, and
$$\delta_{j,\ell}:=(s_{j,\ell},s_{j,\ell+1}].$$
Note that
$\Delta_j$ is the disjoint union of the $\delta_{j,\ell}$'s over $\ell=0,\dots,\ell_j-1$.
Let, finally,

\begin{equation}
B:=\bigcup_{j=j_0}^\infty\bigcup_{q=0}^{\ell_j/2-1}\delta_{j,2q}.
\end{equation}

Take now any large enough natural $n$. Let
\begin{equation}
H_{n,i}:=\Big(\frac i{2^n},\frac{i+1}{2^n}\Big]
\end{equation}
for $i=0,\dots,2^n-1$.

From now on, suppose also that $j\ge n\ge j_0$.

Then $|\delta_{j,\ell}|=1/2^{j+1}<1/2^n=|H_{n,i}|$, where $|\cdot|$ denotes the Lebesgue measure.
For any given $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$, we have
$$B\cap H_{n,i}=\bigcup_{q=q_1}^{q_2}\delta_{j,2q}
\quad\text{and}\quad
H_{n,i}=\bigcup_{\ell=2q_1}^{2q_2+1}\delta_{j,\ell}$$
for some integers $q_1=q_{1;n,i}$ and $q_2=q_{2;n,i}$ such that $0\le q_1\le q_2\le\ell_j/2-1$.
Hence,
$|B\cap H_{n,i}|=\frac12|H_{n,i}|=\frac12\frac1{2^n}$.

So,
\begin{equation}
\frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big)
=\frac14
\end{equation}
-- for each $j\in\{n,n+1,\dots\}$ and each $i=0,\dots,2^n-1$ such that $H_{n,i}\subseteq\Delta_j$.

By $(2)$ and $(1)$, the length $|\Delta_j|$ of the dyadic interval $\Delta_j$ is
$r_{j-1}-r_j\ge\epsilon_{j-1}-\epsilon_j-\tfrac1{2^{j-1}}\ge\tfrac12\,(\epsilon_{j-1}-\epsilon_j)$,
and so, for a given natural $n\ge j_0$, the dyadic interval $\Delta_j$ contains at least
$2^{n-1}(\epsilon_{j-1}-\epsilon_j)$
dyadic intervals $H_{n,i}$.
So, for $a_n$ as defined in the question, we have

\begin{equation}
a_n\ge\frac1{2^n}\sum_{j=n}^\infty\ \sum_{i\colon H_{n,i}\subseteq\Delta_j}
\frac{|B\cap H_{n,i}|}{2^{-n}}\Big(1-\frac{|B\cap H_{n,i}|}{2^{-n}}\Big)
\ge
\frac1{2^n}\sum_{j=n}^\infty 2^{n-1}(\epsilon_{j-1}-\epsilon_j)\frac14
=\frac{\epsilon_{n-1}}8.
\end{equation}
Thus, $a_n\ge\frac{\epsilon_{n-1}}8$ for all $n\ge j_0$.
For instance, letting $\epsilon_j=1/j$ for all natural $j$, we have

$\sum_n a_n=\infty$.