Let $W$ be a Brownian motion and $\sigma$ be a cadlag bounded process independent from $W$. Let $t_{j,n}=j/n$ be the uniform partition of the interval $[0,1]$ and let $\Delta_n=t_{j,n}-t_{j-1,n}=\frac{1}{n}$. It is not difficult to prove that
$$ \sum_{j=1}^{n}\left(\int_{t_{j-1,n}}^{t_{j,n}}\sigma_s\,dW_s\right)^2\stackrel{p}{\longrightarrow}\int_0^1\sigma_s^2\,ds $$
Now I must prove that
$$ \Delta_n^{-1/2}\,\sum_{j=1}^{n}\mathbb{E}_{t_{j-1,n}}\left[\left(\int_{t_{j-1,n}}^{t_{j,n}}\sigma_s\,dW_s\right)^2\left(W_{t_{j,n}}-W_{t_{j-1,n}}\right)\right]\stackrel{p}{\longrightarrow}0 $$
where $\mathbb{E}_{t_{j-1,n}}\left[\cdot\right]$ indicates the $\mathcal{F}_{t_{j-1,n}}$-conditional expected value.
My idea was to use
$$ \int_{t_{j-1,n}}^{t_{j,n}}\sigma_s\,dW_s = \sigma_{t_{j-1,n}}\left(W_{t_{j,n}}-W_{t_{j-1,n}}\right)+O_p\left(\sqrt{\Delta_n}\right), $$
which would imply
\begin{align*} \left(\int_{t_{j-1,n}}^{t_{j,n}}\sigma_s\,dW_s\right)^2\,\left(W_{t_{j,n}}-W_{t_{j-1,n}}\right) &= \sigma_{t_{j-1,n}}^2\left(W_{t_{j,n}}-W_{t_{j-1,n}}\right)^3+\\ &+O_p\left(\Delta_n\right)\,\left(W_{t_{j,n}}-W_{t_{j-1,n}}\right)+\\ &+2\,\sigma_{t_{j-1,n}}\left(W_{t_{j,n}}-W_{t_{j-1,n}}\right)^2\,O_p\left(\sqrt{\Delta_n}\right) \end{align*}
but it is inconclusive.
(This is an extension on my above comment; it does not provide a full answer to the question.)
For right-continuous bounded processes it actually holds that
$$\int_{t_{j-1}}^{t_j} \sigma_s \,d W_s = \sigma_{t_{j-1}} (W_{t_j}-W_{t_{j-1}}) + o_p(\sqrt{\Delta n}); \tag{1}$$
here I use the shorthand $t_j := t_{j,n}$. Indeed: Fix $\epsilon>0$. By Itô's isometry, we have
$$p := \mathbb{P} \left( \frac{1}{\sqrt{\Delta n}} \left| \int_{t_{j-1}}^{t_j} (\sigma_s-\sigma_{t_{j-1}}) \, dW_s \right| > \epsilon \right) \leq \frac{1}{\epsilon^2 \Delta_n} \int_{t_{j-1}}^{t_j} \mathbb{E}((\sigma_s-\sigma_{t_{j-1}})^2) \, ds.$$
Thus,
$$p \leq \frac{1}{\epsilon^2} \mathbb{E} \left( \sup_{s \leq t_{j-1} + \Delta_n} |\sigma_s-\sigma_{t_{j-1}}|^2 \right).$$
Since the process $\sigma$ is bounded and right-continuous, it follows from the dominated convergence theorem that the right-hand side converges to $0$ as $n \to \infty$, and this gives $(1)$.
As far as I can see, $(1)$ should allow you to use your idea to prove the assertion; I haven't checked the details though.