A real monochromatic wave, with bandwidth equal to zero, must be polarized. This is because "bandwidth equal to zero" means that we've sat for infinite time and not observed it to deviate from the standard (polarized, sinusoidal) solution.
Real monochromatic light, on the other hand, can be thought of as an incoherent mixture of wave trains of frequency $\omega$ but finite duration $\tau$, with arbitrary phases and polarizations. This means the phase and polarization will be constant on timescales much shorter than $\tau$ but can be averaged out on longer timescales. However, this makes the light not quite monochromatic: in frequency space, the spectrum broadens out to a smudge of width $\Delta=1/\tau$ about $\omega$.
In practice, $\Delta$ can be made (experimentally!) to be much shorter than $\omega$. For visible light, $\omega\sim10^{-15}\text{ s}$ while the bandwidth can be GHz (for monochromated light) or even down to Hz or smaller (for laser light). This means the light looks like a polarized, definite plane wave for very many periods, so operationally you can make the long-wavelength approximation. It's on timescales longer than $1/\Delta$ that the light looks unpolarized, unless you've taken special precautions. (Some lasers, for example, only emit a single polarization, to very good precision.)
Regarding your second question, it's slightly hard to tell unless you show a bit more working, but it looks like a general fact of the approximation you're making. For any system of size $R$ containing charges oscillating at frequency $\omega$, and not going much further out than $R$ from the system, the limit
$$R\omega/c \ll 1$$
can be seen as the near-field limit (small $R$), the long-wavelength limit (long $\lambda$), the slow-field limit (small $\omega$), or the fast-light limit (changes in the field propagate instantly). Because of the latter, you can solve it quasi-statically: solve the electrostatic problem at each instant and it will give you the electric field. If my memory serves, the change in electric field is usually allowed to induce a magnetic field, but this is not allowed to feed back into the electric field, as it is a factor of $\left(\frac{R\omega}{c}\right)^2$ smaller than the electrostatic contribution.
Your situation does look different, probably because the fast conductive response prohibits this to leading order. You therefore solve the magnetostatic problem at each instant - a conductive sphere in a given constant applied magnetic field - and that gives you the time-dependent magnetic field. This would then be allowed to induce an electric field as $\nabla\times\mathbf{E}=-\partial_t \mathbf B$, but that can't feed back into the magnetic field: the electric field is of the order of $E_0\sim R\omega B_0$, and re-inducing a magnetic field through $\nabla \times \mathbf B=\frac{1}{c^2}\partial_t \mathbf E$ gives a contribution of order $B_1\sim \frac{R \omega}{c^2}E_0\sim\frac{R^2\omega^2}{c^2} B_0\ll B_0$. I think this settles the inequality you observe.
