If $Y(t)$ be a regular adapted process such that $\int_{0}^{t}\mathbb{E}\left[ Y^2(s)\right]ds < \infty$ then
$$\mathbb{E}\left[\int_{0}^{t}Y(s)dW_s\right]=0$$
Set $Y(s)=\sigma(s) e^{iuX(s)} $, we have
$$\int_{0}^{t}\mathbb{E}\left[ Y^2(s)\right]ds=\int_{0}^{t}\mathbb{E}\left[ \sigma^2(s) e^{2iuX(s)}\right]ds=\int_{0}^{t}\sigma^2(s)\mathbb{E} \left[ e^{2iuX(s)}\right]\tag 1$$
note
$$\mathbb{E} \left[ e^{2iuX(s)}\right]=\exp\left(2\text{i}\,u\,\mathbb{E}[X_s]-\frac{1}{2}u^2\text{Var}(X_s)\right)\tag 2$$
X(t) is normally distibuted with zero mean and a variance given by
$$\text{Var}(X_s)=\int_{0}^{s}\sigma^2(u)du$$
thus
$$\int_{0}^{t}\mathbb{E}\left[ Y^2(s)\right]ds=\int_{0}^{t}\sigma^2(s)\exp\left(\int_{0}^{s}\sigma^2(u)du\right)ds<\infty$$