$\int_{-\infty}^{+\infty} |G(f)| \,e^{j2\pi ft}df=|g(t)|$?
The answer is clearly 'no', except for trivial cases. Note that
$$G(f)=|G(f)|e^{j\phi(f)}\tag{1}$$
where $\phi(f)$ is the phase of $G(f)$. Rewriting $(1)$ as
$$|G(f)|=G(f)e^{-j\phi(f)}\tag{2}$$
and taking the inverse Fourier transform gives
$$\tilde{g}(t)=(g\star a)(t)\tag{3}$$
where $\tilde{g}(t)$ is the inverse Fourier transform of $|G(f)|$, and $a(t)$ is the impulse response of a (generally non-causal) all-pass with frequency response $A(f)=e^{-j\phi(f)}$.
If we are given $|G(f)|$, we do not know $\phi(f)$ and, consequently, we don't know $a(t)$, so we can't compute $\tilde{g}(t)$ from $(3)$. Anyway, $(3)$ shows the relation between the inverse Fourier transform of $|G(f)|$ and $g(t)$.
Note that since $|G(f)|$ is real-valued, $\tilde{g}(t)$ must be even
$$\tilde{g}(t)=\tilde{g}(-t)\tag{4}$$