Note that in general the Fourier transform of a stationary process $x(t)$ does not exist. The Wiener-Khinchin theorem only states that under certain conditions the power spectral density of $x(t)$ exists, and it can be computed as the Fourier transform of the autocorrelation function of $x(t)$.
Having said that, if for some reason one assumes that the Fourier transform of $x(t)$ exists, then you can do the math and see if you get a useful result. So, let's see. I assume that $x(t)$ is real-valued. This is not at all necessary, but just simplifies things a bit.
Since $\text{Re}\{X(\omega)\}=\frac12[X(\omega)+X^*(\omega)]$ and $\text{Im}\{X(\omega)\}=\frac{1}{2j}[X(\omega)-X^*(\omega)]$, we can compute the desired expectations from $E\{X(\omega)X(\omega')\}$ and $E\{X^*(\omega)X(\omega')\}$. For the first of these expectations we have
$$E\{X(\omega)X(\omega')\}=E\left\{\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt
\int_{-\infty}^{\infty}x(t')e^{-j\omega' t'}dt' \right\}=\\
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}E\{x(t)x(t')\}e^{-j\omega t}e^{-j\omega't'}dtdt'$$
With the substution $\tau=t'-t$ and with the autocorrelation function $R_x(\tau)=E\{x(t)x(t+\tau)\}$ (assuming wide-sense stationarity of $x(t)$) we obtain
$$E\{X(\omega)X(\omega')\}=\int_{-\infty}^{\infty}R_x(\tau)e^{-j\omega'\tau}d\tau\int_{-\infty}^{\infty}e^{-j(\omega+\omega')t}dt$$
Interpreting the second integral as a distribution, and with the power spectrum $S_x(\omega)=\mathcal{F}\{R_x(\tau)\}$ we finally get
$$E\{X(\omega)X(\omega')\}=2\pi\delta(\omega+\omega')S_x(\omega')=
2\pi\delta(\omega+\omega')S_x(-\omega)=2\pi\delta(\omega+\omega')S_x(\omega)\tag{1}$$
because $S_x(\omega)$ is an even function (we assumed $x(t)$ to be real-valued). The other expectation $E\{X^*(\omega)X(\omega')\}$ can be derived in a completely analogous manner. The result is
$$E\{X^*(\omega)X(\omega')\}=2\pi\delta(\omega-\omega')S_x(\omega)\tag{2}$$
From (1) and (2) we can obtain the desired expectations
$$E\{\text{Re}\{X(\omega)\}\text{Re}\{X(\omega')\}\}=
\pi S_x(\omega)[\delta(\omega-\omega')+\delta(\omega+\omega')]\\
E\{\text{Im}\{X(\omega)\}\text{Im}\{X(\omega')\}\}=
\pi S_x(\omega)[\delta(\omega-\omega')-\delta(\omega+\omega')]$$
It remains to compute the quantities $E\{\text{Re}\{X(\omega)\}$ and $E\{\text{Im}\{X(\omega)\}$. We can easily derive them from the expectations $E\{X(\omega)\}$ and $E\{X^*(\omega)\}$:
$$E\{X(\omega)\}=E\left\{ \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \right\}=
\int_{-\infty}^{\infty}E\{x(t)\}e^{-j\omega t}dt=
\mu_x\int_{-\infty}^{\infty}e^{-j\omega t}dt=2\pi\mu_x\delta(\omega)$$
with $\mu_x=E\{x(t)\}$, where we have again assumed wide-sense stationarity of $x(t)$. Obviously, we get the same result for $E\{X^*(\omega)\}$:
$$E\{X^*(\omega)\}=2\pi\mu_x\delta(\omega)$$
This results in
$$E\{\text{Re}\{X(\omega)\}\}=2\pi\mu_x\delta(\omega)\quad\text{and}\quad
E\{\text{Im}\{X(\omega)\}\}=0$$
So we have
$$E\{\text{Re}\{X(\omega)\}\}\cdot E\{\text{Re}\{X(\omega')\}\}=4\pi^2\mu_x^2\delta(\omega)\delta(\omega')\\
E\{\text{Im}\{X(\omega)\}\}\cdot E\{\text{Im}\{X(\omega')\}\}=0\tag{3}$$
Combining (3) with (1) and (2) you obtain the desired result. From the result you can see that the covariances of the real parts of $X(\omega)$ and $X(\omega')$ vanish everywhere except for $\omega=\pm\omega'$. The same is true for the covariances of the imaginary parts. The difference is an additional term for $\omega=\omega'=0$ for the covariance of the real parts.
