My approach is the following:
we take Fourier transform with respect to $x$, where $k$ is the variable the resulting fourier transform is in.
$\hat{u}_{tt} + k^2 \hat{u} = 0$
Solving this gives me (I think is where I am wrong)
$A(x)e^{ikt}+B(x)e^{-ikt}$
Using the shift rule, I can recover the $F(x-t)$ and $F(x+t)$ but how am I going to get the integral of $G$?
Solving should give you, rather
$$\hat{u}(k,t) = A(k) e^{i k t} + B(k) e^{-i k t}$$
$A$ and $B$ come from initial conditions. You then inverse FT to get $u(x,t)$.
In your case, $u(x,0) = F(x)$ means that, taking the FT
$$\hat{u}(k,0) = A(k) + B(k) = \hat{F}(k)$$
where
$$\hat{F}(k) = \int_{-\infty}^{\infty} dx \: F(x) e^{i k x}$$
The other initial condition is
$$\frac{\partial}{\partial t} \hat{u}(k,0) = i k [A(k) - B(k)] = \hat{G}(k)$$
where
$$\hat{G}(k) = \int_{-\infty}^{\infty} dx \: G(x) e^{i k x}$$
Two equations, two unknowns...
