Note that, by Ito's isometry,
\begin{align*}
E\left(\int_{0}^{t} F(s)\,dW_{1}(s)\int_{0}^{t} G(s)\, dW_{2}(s)\right) &= \int_0^t E\big(F(s) G(s) \big) d\langle W_1, W_2\rangle_t =0.
\end{align*}
That is, $\int_{0}^{t} F(s)\,dW_{1}(s)$ and $\int_{0}^{t} G(s)\,dW_2(s)$ are uncorrelated.
If $F(s)$ and $G(s)$ are deterministic, then, for any constants $a$ and $b$
\begin{align*}
a\int_{0}^{t} F(s)\,dW_{1}(s) + b\int_{0}^{t} G(s)\,dW_2(s)
\end{align*}
is the limit, in probability, of the sum
\begin{align*}
\sum_{i=1}^n \Big(aF(t_{i-1}) \big(W_1(t_i)-W_1(t_{i-1})\big) +bG(t_{i-1}) \big(W_2(t_i)-W_2(t_{i-1})\big)\Big),\tag{1}
\end{align*}
where $0=t_0 < t_1 < \cdots < t_n = t$ is a partition of $[0, t]$. Note that all terms in $(1)$ are normal and independent, and then their sum is also normal. Since the limit, in probability, of normal random variables is also normal, \begin{align*}
a\int_{0}^{t} F(s)\,dW_{1}(s) + b\int_{0}^{t} G(s)\,dW_2(s)
\end{align*}
is normal. That is,
$\int_{0}^{t} F(s)\,dW_{1}(s)$ and $\int_{0}^{t} G(s)\,dW_2(s)$ are jointly normal, and are therefore independent (note that the joint characteristic function is the product of the individual characteristic functions).
However, if $F(s)$ and $G(s)$ are not deterministic, then $\int_{0}^{t} F(s)\,dW_{1}(s)$ and $\int_{0}^{t} G(s)\,dW_2(s)$ are not necessarily independent. For example, consider $X_t=\int_{0}^{t} W_2(s)\,dW_{1}(s)$ and $Y_t=\int_{0}^{t} W_1(s)\,dW_2(s)$. Then $X_t$ and $Y_t$ is uncorrelated, but are not independent.
$X_t$ and $Y_t$ are uncorrelated but are not independent.
Note that
\begin{align*}
dX_t^2 = 2X_tdX_t + W_2^2(t) dt,
\end{align*}
and
\begin{align*}
dY_t^2 = 2Y_t dY_t + W_1^2(t) dt.
\end{align*}
Then,
\begin{align*}
X_t^2 Y_t^2 = 2\int_0^t Y_s^2 X_s dX_s+2\int_0^t X_s^2 Y_s dY_s + \int_0^t \left(W_1^2(s)X_s^2 + W_2^2(s) Y_s^2 \right) ds.
\end{align*}
By symmetry,
\begin{align*}
E\left( X_t^2 Y_t^2\right) = 2\int_0^t E\left(W_1^2(s)X_s^2\right) ds.\tag{2}
\end{align*}
Note that
\begin{align*}
dW_1^2(t) = 2W_1(t) dW_1(t) + dt.
\end{align*}
Then
\begin{align*}
W_1^2(t)X_t^2 &= 2\int_0^t X_s W_1(s)^2 dX_s + 2\int_0^t W_1(s) X_s^2 dW_1(s)\\
&\qquad +\int_0^t \left(W_1(s)^2 W_2(s)^2 + X_s^2 + 4X_sW_1(s)W_2(s)\right)ds.\tag{3}
\end{align*}
Furthermore, note that
\begin{align*}
W_1(t)W_2(t) = X_t + Y_t.
\end{align*}
Then,
\begin{align*}
E\left( X_tW_1(t)W_2(t)\right) &= E\left(X_t^2 + X_t Y_t\right)\\
&=\int_0^t s ds=\frac{1}{2}t^2.
\end{align*}
Therefore, from $(3)$,
\begin{align*}
E\left(W_1^2(t)X_t^2\right) &=\int_0^t \left(s^2 + \frac{1}{2}s^2 + 2s^2 \right)ds = \frac{7}{6}t^3,
\end{align*}
and, from $(2)$,
\begin{align*}
E\left( X_t^2 Y_t^2\right) &= 2\int_0^t E\left(W_1^2(s)X_s^2\right) ds\\
&=\frac{7}{3} \int_0^t s^3ds = \frac{7}{12} t^4.
\end{align*}
However,
\begin{align*}
E\left( X_t^2 \right)E\left(Y_t^2\right) = \frac{1}{4}t^4.
\end{align*}
That is, $X_t$ and $Y_t$ are not independent.
