Plancherel's theorem states that the Fourier transform preserves the $L^2$ norm, meaning that if $f$ and $g$ are functions whose Fourier transforms are denoted by $\hat{f}$ and $\hat{g}$ respectively, then:
$$ \| f \|_{L^2} = \| \hat{f} \|_{L^2} $$
Similarly, is it true that: if $\mu_1$ and $\mu_2$ are probability measures and $\hat{\mu}_1$ and $\hat{\mu}_2$ are their respective Fourier transforms, then the $L^2$ distance between $\hat{\mu}_1$ and $\hat{\mu}_2$ is equal to the Wasserstein-2 distance between $\mu_1$ and $\mu_2$?
Mathematically, this can be expressed as:
$$ W_2(\mu_1, \mu_2) = \left( \int_{X \times X} |x - y|^2 d\pi(x,y) \right)^{1/2} = \left( \int_{\mathbb{R}} |\hat{\mu}_1(\xi) - \hat{\mu}_2(\xi)|^2 d\xi \right)^{1/2} $$
No, this is not true, why would it be ? In particular your second distance is nothing but the $L^2$ norm of $\mu_1-\mu_2$, which is not even always defined when $\mu_1$ and $\mu_2$ are not absolutely continuous with respect to the Lebesgue measure.
If you want a counterexample, take $\mu_1 = \delta_0$ and $\mu_2 = \delta_a$, then their Wasserstein distance is $|a|$, but $$ |\hat{\mu}_1(\xi)-\hat{\mu}_2(\xi)| = |1-e^{2i\pi\,a\,\xi}| = 2\left|\sin(\pi\,a\,\xi)\right| $$ which is clearly not square integrable ... so the integral is infinite.
Actually, the Wasserstein distance should rather be compared with negative Sobolev distances, such as the $H^{-1}$ norm.