= Solution
Put $R_\varepsilon=\int_0^\varepsilon(H_u-H_0)\,dB_u$ and $d_\varepsilon=\varepsilon^{-1}\int_0^\varepsilon\mathbb E|H_u-H_0|^2du$. Boundedness and path continuity of $H$ make $\mathbb E|H_u-H_0|^2\to0$ by the <dominated convergence theorem>, so $d_\varepsilon\to0$. The <Itô isometry> gives $\mathbb E R_\varepsilon^2=\varepsilon d_\varepsilon$.
We must not assume independence between the integral and its Brownian denominator. Instead, the <Cauchy-Schwarz inequality> gives
$$
\mathbb E\left|\frac{R_\varepsilon}{B_\varepsilon}\right|^{1/4}
\leq\bigl(\mathbb E|R_\varepsilon|^{1/2}\bigr)^{1/2}\bigl(\mathbb E|B_\varepsilon|^{-1/2}\bigr)^{1/2}.
$$
Concavity, or <Jensen inequality>, bounds the first factor by $(\mathbb E R_\varepsilon^2)^{1/8}$. Since $B_\varepsilon\overset d=\sqrt\varepsilon N$, the second factor is $\varepsilon^{-1/8}(\mathbb E|N|^{-1/2})^{1/2}$. This negative normal moment is finite: the density is bounded near zero, where $\int_0^1u^{-1/2}du<\infty$, and has Gaussian decay at infinity. In fact $\mathbb E|N|^{-1/2}=2^{-1/4}\Gamma(1/4)/\sqrt\pi$. The powers of $\varepsilon$ cancel, yielding the <fractional-moment control of Brownian increment ratios>
$$
\boxed{\mathbb E\left|\frac{R_\varepsilon}{B_\varepsilon}\right|^{1/4}\leq C d_\varepsilon^{1/8}\longrightarrow0,\qquad C=(\mathbb E|N|^{-1/2})^{1/2}.}
$$
The ratio may be assigned any value on the null event $B_\varepsilon=0$.
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