Eigenenergies in rotating frame From Sakurai, unitary equivalent observables have identical spectra.
$$ A |a\rangle = a |a\rangle, $$
$$ UAU^{-1}U|a\rangle = aU|a\rangle $$
Which can be written as,
$$ B|b\rangle = a |b\rangle $$
where, $$ B = UAU^{-1}$$ 
However, if I'm not wrong, this works when the unitary transformation U is time-independent. What if the unitary transformation depends on time, as in the rotating frame. Then, the new Hamiltonian B should be
$$ B = UAU^{-1} + i \hbar \dot{U} U^{-1} $$
From this, the eigenergies should definitely change. Am I doing anything wrong? Does it mean there will be different physics in a rotating frame? 
 A: You are correct that the eigen-energies change in the new reference frame. But that doesn't mean that there are different physics in that frame. 
This is not at all restricted to quantum mechanics. For example, consider the simple classical problem of a spinning top. In the "lab" frame, the top has a kinetic energy $I \omega^2/2$, but in a (non-inertial) reference frame rotating along with the top, the top seems stationary, i.e. the kinetic energy seems to be zero. Nevertheless, you can still describe the dynamics in this rotating frame by taking non-inertial forces into account.
You can even think of a more simpler classical problem, just a particle moving at a constant velocity with respect to the lab frame has non-zero kinetic energy, while the same particle has zero kinetic energy in a frame that's moving alongside it with the same velocity.
The same is true for the quantum case, where the analog of the non-inertial forces in classical mechanics is exactly the $i \hbar \dot{U}U^{-1}$ you're referring to (that term is actually sometimes called the "inertial term").
In practice, the rotating frame is usually taken into account using quadrature detection. For example, in magnetic resonance (MR), your lab-frame Hamiltonian is something like 
$$H(t) = \omega_0 S_z + \omega_1 \big(\cos(\omega t) S_x - \sin(\omega t) S_y\big),$$
which, taken to the frame rotating around $z$ with rate $\omega$, transforms to (using $U = e^{-i \omega t S_z}$)
$$\tilde{H} = (\omega_0-\omega)S_z + \omega_1 S_x.$$
Now the observable that is measured in MR is typically the transverse magnetization. If I call the axes of the rotating frame $\tilde{x},\tilde{y},\tilde{z}$, and the lab frame $x,y,z$, the lab-frame observable that is actually measured, is
$$\langle S_x \rangle = \langle \tilde\psi(t) \vert S_x \vert \tilde\psi(t) \rangle = \langle \psi(t) \vert e^{i \omega t S_z} S_x e^{-i \omega t S_z} \vert \psi(t) \rangle,$$
or,
$$\langle S_x \rangle = \langle \psi(t) \vert \big(\cos(\omega t) S_x - \sin(\omega t) S_y\big) \vert \psi(t) \rangle,$$
meaning that the magnetization components of the lab and rotating frames are related by
$$\langle S_x \rangle = \cos (\omega t) \langle S_{\tilde x} \rangle - \sin (\omega t) \langle S_{\tilde y} \rangle.$$
Now observe what happens if I multiply this equation by $\cos(\omega t)$:
$$\langle S_x \rangle \cos(\omega t) = \cos^2(\omega t) \langle S_{\tilde x} \rangle - \cos (\omega t)\sin (\omega t) \langle S_{\tilde y} \rangle = \frac{1 + \cos(2 \omega t)}{2}\langle S_{\tilde x} \rangle - \frac{\sin(2 \omega t)}{2}\langle S_{\tilde y} \rangle,$$
so if we pass this through a low-pass filter, we get
$$\frac{1}{2} \langle S_\tilde{x} \rangle.$$
And you can do the same to get $S_\tilde{y}$. 
So to summarize, what happens in actual experiment is that you directly measure the lab-frame observable $S_x$, and then transform that into a rotating frame observable by a combination of mixing with the right carrier/ low-pass filtering.
This process is typically done automatically in things like MR spectrometers, such that the experimenter doesn't even have to worry about a lab frame anymore. The data shown to him by the device have already been transformed to the rotating frame, so there is no need to use the lab frame for anything, making things a lot easier.
In practice, this process is a bit more complicated, but that is the idea. You can read more about it in, e.g., here.
