= Solution
Define $\Omega=\sqrt{\Delta^2+\Omega_R^2}$. Squaring the two-by-two matrix gives $H_R^2=\hbar^2\Omega^2 I/4$, so
$$
\boxed{E_\pm=\pm\frac{\hbar\Omega}{2},\qquad
|\Psi_\pm(t)\rangle=e^{\mp i\Omega t/2}e^{-i\omega tS_z/\hbar}|\pm\rangle_R}.
$$
Here $|\pm\rangle_R$ is the corresponding normalized <eigenstate> at $t=0$. In the rotating frame only its <quantum phase> changes. In the laboratory frame its <spin> direction also rotates about $z$ at <angular velocity> $\omega$: $\langle\mathbf S\rangle_\pm=\pm\hbar(\Omega_R\cos\omega t,\Omega_R\sin\omega t,\Delta)/(2\Omega)$. An arbitrary initial state is the same fixed linear combination of these two solutions. If $\Omega=0$, the rotating-frame <Hamiltonian operator> vanishes and the entire doublet is degenerate; evolution is just $R(t)$.
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