Integral $I_t$ does not seem to follow the normal distribution.
$\newcommand{\d}{\mathrm{d}}$Define $I_t = \int_0^{t} W_{\sqrt{u}}^2 \d u$ and let $u = v^2$ as you have already suggested.
By integration by parts
$$
I_t = \int_0^{\sqrt{t}}W_v^2 \d v^2 = \underbrace{W_v^2 v^2\big|_0^{\sqrt{t}}}_{tW_{\sqrt{t}}^2} - \underbrace{\int_0^{\sqrt{t}}v^2\d W_v^2}_{J_t},
$$
where $\d W_v^2 = 2W_v \d W_v + \d v$, therefore,
$$
J_t = \underbrace{\int_0^{\sqrt{t}} 2W_v v^2 \d W_v}_{K_t} + \underbrace{\int_0^{\sqrt{t}}v^2 \d v}_{\tfrac{1}{3}t^{3/2}}
$$
Now define $f(v, x) = vx^2$ and apply Itô's formula:
\begin{align}
&\d(v W_v^2) {}={} W_v^2 \d v + 2vW_v \d W_v + \tfrac{1}{2}(2v)\d v\\
\Rightarrow{}& \sqrt{t}W_{\sqrt{t}}^2{}={}\int_0^{\sqrt{t}} W_v^2 \d v + K_t + \int_0^{\sqrt{t}}v\d v\\
\Rightarrow{}& K_t = \sqrt{t}W_{\sqrt{t}}^2-\tfrac{1}{2}t - \int_0^{\sqrt{t}} W_v^2 \d v,
\end{align}
therefore
\begin{align}
I_t {}={}& tW_{\sqrt{t}}^2 - K_t - \tfrac{1}{3}t^{3/2}
\\
{}={}&(t - \sqrt{t})W_{\sqrt{t}}^2
+ \tfrac{1}{2}t
- \tfrac{1}{3}t^{3/2}
+ \int_0^{\sqrt{t}} W_v^2 \d v
\end{align}
Note that $W_{\sqrt{t}}^2$ follows a chi-squared distribution, the second and third terms are deterministic and the integrands of the last term, $W_v^2$ follow a chi-squared distribution too.
Regarding the last integral, you can check out this question on QSE.
