Note: I'm using $$S_F(\omega) \equiv \int_{-\infty}^\infty dt e^{i \omega t} \left \langle F(0)F(t) \right \rangle$$
because I believe it is considerably more standard than the definition given in the question post.
What physical information does the power spectrum $S_F(\omega)$ provide about the nature of the random variable $F(t)$ in this case?
The power spectral density tells you exactly what it sounds like: the amount of fluctuations at a given frequency.
There are a few ways to see the meaning of "amount of fluctuations at a given frequency", the most direct and possibly intuitive is to directly relate how much the particle moves around to its power spectral density.
Mean square motion of particle
Let us ask a concrete physical question: given that the particle starts at $x=0$, what is it the probability distribution of its position after a time $t$?
Since the motion of the particle is presumably due to a sum of a large number of small fluctuations, the central limit theorem guarantees that the probability distribution is Gaussian.
Therefore, it's enough for us to compute the variance.
We can always write $x(t)$ as an integral of its velocity, i.e.
$$x(t) = \int_0^t \dot{x}(t') \, dt' \,.$$
Therefore,
\begin{align}
\left \langle x(t)^2 \right \rangle
&= \int_0^t dt' \int_0^t dt'' \left \langle \dot{x}(t') \dot{x}(t'') \right \rangle \\
&= \int_0^t dt' \int_0^t dt'' \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_{\dot{x}}(\omega) e^{-i \omega (t' - t'')} \\
&= \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_{\dot{x}}(\omega) \int_0^t dt' \int_0^t dt'' e^{-i \omega (t' - t'')} \\
&= 2 t^2 \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_{\dot{x}} (\omega) \underbrace{\left( \frac{1 - \cos(\omega t)}{(\omega t)^2} \right)}_{W(\omega,t)} \, . \tag{$\star$}
\end{align}
This result can be interpreted as a weighting function $W(\omega, t)$ integrated against the spectral density.
Figure 1: $W(\omega, t)$ for the free particle case.
The exact form of this weighting function is particular to the case where we let the particle drift freely.
If we were to filter the particle's motion (e.g. if this were an electrical signal instead of a physical particle) or change the boundary conditions in some way, we'd have a different form for $W$.
In the case of strong damping, which I believe is the case for Brownian motion, we have a large drag force $F_\text{drag} = - \gamma \dot{x}$ and Newton's law goes like this
\begin{align}
F_\text{total} &= m\ddot{x} \\
F_\text{external} - \gamma \dot{x} &= m\ddot{x} \leftarrow \text{(small compared to other side)} \\
F_\text{external}/\gamma &= \dot{x} \, .
\end{align}
Therefore, force and velocity are proportional to each other.
The $W$ we got in this case is large for small $\omega$.
This means that the diffusion of the free particle is dominated by force fluctuations at low frequency.
This isn't surprising: if the spectral density is large at low frequency then the force has a large roughly constant component, which pushes the particle in one direction.
Spectral density at higher frequencies corresponds to force fluctuations that oscillate in time, and contribute less to the particle's total movement in a single direction.
Power spectral density as "power per bandwidth"
This section is a work in progress.
I've saved it in an incomplete state so that I can work on it from another computer.
At the beginning of this post, we said that the power spectral density tells you the amount of fluctuation at a given frequency.
In fact, the power spectral density tells us the power per frequency bandwidth.
Let's see what that means quantitatively.
First, rewrite $(\star)$ with the variable change $z \equiv \omega t$:
$$\langle x(t)^2 \rangle
=2 t \int_{-\infty}^\infty \frac{dz}{2 \pi} S_{\dot{x}}(z/t)
\underbrace{\left( \frac{1 - \cos(z)}{z^2} \right)}_{W(z)} \, .$$
If $S_\dot{x}(z/t)$ is a constant $S_\dot{x}$ over the range where $W(z)$ has substantial weight, then we can factor $S$ out of the integral, do the integral, and arrive at
\begin{align}
\langle x(t)^2 \rangle &= t S_\dot{x} \\
\text{or} \quad \langle x(t)^2 \rangle / t &= S_\dot{x} \, .
\end{align}
Thinking of $\langle x(t)^2 \rangle$ as "energy", the left hand side is energy per time or power.
Therefore, $S_\dot{x}$ is the
