Weak convergence of stochastic integral Consider a sequence of processes $Z_t^n$ and a procoss $Z_t$, $t\in[0,1]$ such that all $\int_0^1 Z^n dW$ and $\int_0^1 Z dW$ are martingales. Assume
$$\int_0^1 Z_t^n \mathrm dW_t \xrightarrow{d} \int_0^1 Z_t \mathrm dW_t$$
in distribution. Do we have
$$\int_0^1 (Z_t^n)^2 \mathrm dt \xrightarrow{d} \int_0^1 Z_t^2 \mathrm dt$$
in distribution? Intuitively, this should be true as the quadratic variation is just a continuous functional of the martingale. But I do not know, how to prove this. Do you have any hints for me? In case the statement is false, under which conditions do we have weak convergence of the quadratic variation?
Moreover, I would be interested in the following extensions of the question: Does such a result hold for general martingales converging in distribution? Does the converse also hold true?
 A: The answer to your first question is negative. 
The best way to answer a question is to pose it in an appropriate way.
So let us address a simpler and more natural question:

Assume that 
  $$
\int_0^1 Z_t dW_t \overset{d}{=} \int_0^1 Y_t dW_t.
$$
  Is it true that 
  $$
\int_0^1 Z_t^2 dt \overset{d}{=} \int_0^1 Y_t^2 dt?\tag{1}
$$

It is clear that if the answer to this question is negative, then it is so for your question as well. 
Now take $Z_t \equiv 1$ so that $\int_0^1 Z_t dW_t = W_1$ is standard Gaussian. For a process $Y$, equality (1) means that $Y$ is deterministic. Therefore, in order to construct a counterexample to this statement, we should find a non-deterministic process $Y$ such that $\int_0^1 Y_t^2 dt$ is standard Gaussian. In other words, we should find a standard normal random variable such that the integrand in its Itô representation is non-deterministic. And this is very easy to construct: 
$$
\xi = f(W_1) \text{ with } f(x) = x\left(\mathbf{1}_{[-1,1]}(x)- \mathbf{1}_{\mathbb R\setminus [-1,1]}(x)\right).
$$
It is clear that $\xi\simeq N(0,1)$. But the integrand $Y$ in 
$$
\xi = \int_0^1 Y_t dW_t
$$
is non-deterministic. Indeed, if it were the case, the random variable
$$
\xi_s = \mathsf E\left[\xi\mid \mathcal F^W_s\right] = \mathsf E\left[f(W_s + (W_1 - W_s)\mid \mathcal F_s^W\right] = \mathsf E\left[f(x + (W_1 - W_s))\right]\big|_{x = W_s}
$$
would have a Gaussian distribution, but it does not. 
Alternatively, one can use the Clark formula: $Y_s = \mathsf E[D_s f(W_1)\mid \mathcal F^W_s]$, where $D_s$ is the Malliavin derivative. Using the chain rule, 
$$
Y_s = \mathsf E[ f'(W_1) D_s W_1\mid \mathcal F^W_s]  = \mathsf E \left[f'(W_s + (W_1 - W_s))\mid \mathcal F^W_s\right]
\\ = \mathsf E\left[f'(x + (W_1 - W_s))\right]\big|_{x = W_s} = g(W_s),
$$
where $f'(x) = \mathbf{1}_{[-1,1]}(x) - \mathbf{1}_{\mathbb R\setminus [-1,1]}(x)$. It is easy to check that $g$ is not constant.

Seems that under the assumption that
$$
\int_0^\cdot Z^{(n)}_t dW_t \overset{d}{\longrightarrow} \int_0^\cdot Z_t dW_t
$$
as processes,
$$
\int_0^\cdot \left(Z^{(n)}_t\right)^2 dt\overset{d}{\longrightarrow} \int_0^\cdot Z^2_t dt 
$$
as processes. Write if you are interested in this, then I will consult the books I have.
