Hint Let $\varepsilon>0$, $\delta>0$. We have
$$\begin{align*} \mathbb{P} &\left( \left| \frac{1}{B_{\varepsilon}} \cdot \int_0^{\varepsilon} H_s \, dB_s - H_0 \right|>\delta \right) \\ &= \mathbb{P} \left( \left| \frac{1}{B_{\varepsilon}} \cdot \int_0^{\varepsilon} (H_s-H_0) \, dB_s \right|>\delta \right) \\
&\leq \mathbb{P} \left( \left| \frac{1}{B_{\varepsilon}} \cdot \int_0^{\varepsilon} (H_s-H_0) \, dB_s \right|>\delta, \left| \frac{\sqrt{\varepsilon}}{B_{\varepsilon}} \right| \leq K \right)+ \mathbb{P} \left( \left| \frac{\sqrt{\varepsilon}}{B_{\varepsilon}} \right| > K \right) \\
&\leq \mathbb{P} \left( \left| \frac{1}{\sqrt{\varepsilon}} \cdot \int_0^{\varepsilon} (H_s-H_0) \, dB_s \right|>\frac{\delta}{K} \right)+ \mathbb{P} \left( \frac{|B_{\varepsilon}|}{\sqrt{\varepsilon}} < \frac{1}{K} \right)\\
&=: I_1+I_2 \end{align*}$$
for any $K>0$. Since $\frac{B_{\varepsilon}}{\sqrt{\varepsilon}} \sim N(0,1)$, we can choose $K>0$ (independent of $\varepsilon$) such that
$$I_2 \leq \frac{\varepsilon}{4}$$
For the first term $I_1$ apply Markov's inequality and Itô's isometry to show that it converges to zero as $\varepsilon \to 0$, using the continuity of $H$ at $0$.
Remark A detailed proof can be found in Dean Isaacson, Stochastic Integrals and Derivatives (1969).
I answer in the case of dimension 1.
Call $(L_t^x)_{t \ge 0}^{x \in \mathbb{R}}$ the local time process of $W$ and $(\tau_r)_{r \ge 0}$ the inverse of local time at $0$.
Then, by Markov property, the process $((L_{\tau_r}^x)^{x \in \mathbb{R}})_{r \ge 0}$ (with values in the space of all non-negative continuous functions on $\mathbb{R}$ with compact support) have independent and stationary increments.
Moreover, $\mathbb{E}[L_{\tau_r}^x] = r$ for all $r$ and $x$ (it follows from Ray - Knight theorem, but it can also be proved directly with stochastic calculus). Integrating with regard to $x$, and using
$$O_A(t) = \int_A 1_{A}(x) L_t^x dx,$$
we get that for every Borel subset $A$ of $\mathbb{R}$,
$\mathbb{E}[O_A(\tau_r)] = r\lambda(A)$.
If $A_1$ and $A_2$ are Borel subsets with the same finite Lebesgue measure, such that $A_1 \triangle A_2$ has positive Lebesgue measure, the
the process $(O_{A_1}(\tau_r)-O_{A_2}(\tau_r))_{r \ge 0}$ is centered, not deterministic, and has independent and stationary increments.
Hence this process cannot be bounded, although law of large numbers applies
and central limit theorem also (the random variables are square integrable).
Therefore, the process $(O_{A_1}(t)-O_{A_2}(t))_{t \ge 0}$ cannot be bounded.
EDIT : getting an upper bound of moments at any stopping time $\tau$, including constant times.
If $A_1$ and $A_2$ are Borel subsets with the same finite Lebesgue measure, $O_{A_1}(\tau)-O_{A_2}(\tau)$ can be written as this Lebesgue measure times the integral of $L_\tau^x-L_\tau^y$ with regard to $d\pi(x,y)$, where $\pi$ is any coupling of the uniform measures on $A_1$ and $A_2$. Hence it suffices to bound above the moments of $L_\tau^x-L_\tau^y$.
By Tanaka's formula, for $t \ge 0$,
$$L_t^x-L_t^y = |B_t-x|-|B_t-y|-|x|+|y| + M^{x,y}_t,$$
where $M^{x,y}$ is the martingale defined by
$$M^{x,y}_t = \int_0^t (\mathrm{sign}(B_s-x)-\mathrm{sign}(B_s-y)) dB_s.$$
Thus for every stopping time $\tau$ (including constant times)
$$|L_\tau^x-L_\tau^y| \le 2|x-y| + |M^{x,y}_\tau|.$$
Given $p \ge 1$, we derive
$$\Vert L_\tau^x-L_\tau^y \Vert_p^p \le \Big( 2|x-y| + \Vert M^{x,y}_\tau \Vert_p\Big)^p.$$
To get an upper bound, we use Barlow - Yor inequalities.
$$\Vert M^{x,y}_\tau \Vert_p^p \le C_p\mathbb{E}[\langle (M^{x,y}_\tau)^{p/2} \rangle],$$
where $C_p$ is some positive constant.
Note that its quadratic variation is given by
$$\langle M^{x,y} \rangle_t
= \int_0^t (\mathrm{sign}(B_s-x)-\mathrm{sign}(B_s-y))^2 ds
= 4\int_0^t \mathbb{1}_{[\min(x,y) \le B_s \le \max(x,y)]} ds,$$
since
$$\int_0^t \mathbb{1}_{[B_s \in \{x,y\}} ds = 0.$$
A trivial upper bound is $\langle M^{x,y} \rangle_t \le 4t$.
A less trivial upper bound is
$$\mathbb{E}[\langle M^{x,y} \rangle_t]
= \int_0^t \mathbb{P}[B_s \in \{x,y\}] ds
\le \int_0^t \frac{1}{\sqrt{2\pi s}} |x-y| ds
= \frac{\sqrt{2s}}{\sqrt{\pi}} |x-y|.$$
Best Answer
The (one-dimensional) Brownian motion admits "local times" (occupation-time densities) $L^x_t(\omega)$, adapted to the Brownian filtration, jointly continuous in $(x,t)$ (a.s.), such that $$ \int_0^t g(X_s)\,ds =\int_{\Bbb R} g(x) L^x_t\,dx, \qquad\forall t\ge 0, $$ almost surely. Here $g$ is a Borel function that is non-negative or bounded (for example). With this fact in hand, the answer to your question should be clear.