With circularly polarised waves, the effect can only be observed interferometrically, because the Faraday rotation simply becomes the imposition of different phase delays on the two circularly polarised components. To understand this, witness that the Faraday rotation on a general polarisation state expressed with linear polarisation state basis is:
$$\left(\begin{array}{c}x\\y\end{array}\right)\mapsto \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$
and then we transform to circular polarisation basis states $f_+,\,f_-$ by:
$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)= \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$
so that our Faraday rotation through angle $\theta$ above becomes:
$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)\mapsto \left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)^{-1}\left(\begin{array}{cc}f_+\\f_-\end{array}\right) = \left(\begin{array}{cc}e^{i\,\theta}&0\\0&e^{-i\,\theta}\end{array}\right)\left(\begin{array}{cc}f_+\\f_-\end{array}\right)$$
Indeed, this is the essence of the spin-1 particle: its eigenstates pick up a phase of $e^{i\,\theta}$ on rotation through angle $\theta$ (unlike an electron, which would pick up a phase of $e^{i\,\frac{\theta}{2}}$).
Edit: Your understanding expressed in your question seems to be right. In the above I give the Jones matrix, i.e. the linear transformation wrought on the vector of superposition weights, for the two cases where (1) linear polarisation states and (2) circular polarisation states are used as the state space basis: in the former case, the effect is a rotation of the plane of polarisation through $\theta$ radians, in the latter the right handed circular polarised component's phase is increased by $\theta$ radians, the left hand component's phase is decreased by $\theta$ radians. In all cases, this $\theta$ is given by $\theta = V\, B_\parallel \,L$ where $L$ is the crystal's length, $B_\parallel$ the magnetic induction component along the optical axis and $L$ the crystal's length.
Edit 2: Well, the results for circularly polarised loght follow from: (1) experimental measurements for linearly polarised light and (2) experimental confirmation of the system's linearity; once we have (2) then circular / linear description is simply a change of basis. Moreover, you can do the experiments for circularly polarised light, you're simply looking for changes in phase. So you can do the experiment interferometrically. Or, you can do an experiment as below with waveplates and the rotator.
The rotator works exactly the same with linearly polarised input light whether or not the waveplates are in place, confirming the linearity of the system. So it is acting on the circularly polarised light states exactly as my basis trasnformation equations above foretell. At a deeper level, the waveplate's action as a transformer to and from circularly polarised light can be checked by the following experiment in 1936 by Richard Beth:
Richard A. Beth, "Mechanical Detection and Measurement of the Angular Momentum of Light" Phys. Rev. 50 July 15 1936
Best Answer
A waveplate is a birefringent crystal. Birefringence is a particular kind of anisotropy where the refractive index depends on the plane of polarization of the light: there are two, orthogonal linear polarization planes which have a relatively lower ("fast axis") and higher ("slow axis") refractive index. An waveplate of angle $\phi$ is of such a thickness that the phase delay differs for these two polarization states by $\phi$ as light passes through the crystal.
Therefore, the way to analyze the action of such a crystal is to translate the above into equations. We represent a general polarization state by a $2\times2$ vector:
$$\left(\begin{array}{c}\alpha\,\cos(\omega\,t+\delta)\\\beta\,\cos(\omega\,t+\epsilon)\end{array}\right)$$
where $\alpha,\,\beta,\,\delta,\,\epsilon$ encode the magnitude and phase of the component of the electric field vector along the fast and slow axes.
In a circular polarization state, $\beta = \alpha = 1$ and $\epsilon = \delta\pm \pi/2$, which means that the fast polarization state leads/ lags the slow by $\pi/2$. This in turn means that if the fast polarization amplitude oscillates as $\cos\omega\,t$, the slow oscillates by $\sin\omega\,t$. It's not hard to show that the head of the vector:
$$\left(\begin{array}{c}\cos(\omega\,t)\\\cos(\omega\,t\pm\frac{\pi}{2})\end{array}\right)=\left(\begin{array}{c}\cos(\omega\,t)\\\mp\sin(\omega\,t)\end{array}\right)$$
undergoes uniform circular motion and that this is therefore circular polarization.
If such a state is input to a quarter wave plate, then the plate imparts a further $\pi/2$ phase delay between the states. So now our state becomes
$$\left(\begin{array}{c}\cos(\omega\,t)\\\cos(\omega\,t\pm\frac{\pi}{2}\pm\frac{\pi}{2})\end{array}\right)=\left(\begin{array}{c}\cos(\omega\,t)\\\pm\cos(\omega\,t)\end{array}\right)$$
and the fast and slow components are now either in or exactly out of phase. It's not hard to see that the head of this vector undergoes simple harmonic rectillinear motion, and that therefore we're now dealing with linear polarization.