Your question is ill-posed.
Basically, when you say that you want to consider
a radiation field described by a quantum coherent state $|\alpha\rangle$
you are already implicitly considering the polarization of the electric field. This is because to speak of a quantum coherent state $|\alpha\rangle$ over a single mode you are already implicitly basing your treatment on a separation of the electromagnetic field observables over a basis of classical modes, i.e. on an expansion of the form
$$
{\mathbf E}(\mathbf r,t) = \sum_n\bigg[a_n(t)\mathbf f_n(\mathbf r)+a_n(t)^*\mathbf f_n^*(\mathbf r)\bigg],
$$
where $\mathbf f_n(\mathbf r)$ are vector-valued functions of position and the (complex) mode amplitudes $a_n(t)$ carry all of the dynamical information about the classical state of the field, and which you then translate via canonical quantization to an expansion of the form
$$
\hat{\mathbf E}(\mathbf r) = \sum_n\bigg[\hat{a}_n\mathbf f_n(\mathbf r)+\hat {a}_n^\dagger\mathbf f_n^*(\mathbf r)\bigg],
$$
where the classical mode amplitudes $a_n(t)$ get replaced by bosonic annihilation operators $\hat a_n$. (A previous appearance of this description on answer to a question of yours is here.)
Generically speaking, moreover, saying that the radiation is described by a coherent state $|\alpha\rangle$ implicitly implies that it is one of those modes that carries the coherent state in question, and that all the other modes are in the QED vacuum state.
Thus,
- is the electric field polarized? yes, by implicit assumption.
- what polarization does it have? it depends ─ whatever polarization is defined by the mode function $\mathbf f_n(\mathbf r)$.
Moreover, that polarization can be anything: it can be linearly polarized, it can be circularly polarized, and it can also be a space-dependent polarization like a radial or azimuthal polarization or more complicated combinations like, say, ellipse-point polarization fields.
On the other hand all of this can, of course, be quantified: in the Schrödinger picture we normally have the time dependence $t\mapsto |\alpha e^{-i\omega t}\rangle$, where the frequency $\omega$ comes from demanding that $\mathbf f_n(\mathbf r)$ be a divergenceless Helmholtz eigenfunction, and this will then give the time dependence
$$
\langle \hat{\mathbf E}(\mathbf r) \rangle
=
\alpha e^{-i\omega t}\mathbf f_n(\mathbf r)+\alpha^* e^{+i\omega t}\mathbf f_n^*(\mathbf r).
$$
If your mode was linearly polarized to begin with then you can set e.g. $\mathbf f_n(\mathbf r) = \mathbf e_x e^{ikx}$, or if you wanted a circular polarization you could swap the $\mathbf e_x$ for $\mathbf e_x+ i \mathbf e_y$, or you could have a space-dependent polarization. In the 'worst'-case scenario, $\mathbf f_n(\mathbf r)$ could be a space-dependent polarization field that varies on length scales much shorter than what your detector can resolve, in which case the field will look de-polarized to your detector.
However:
it is important to note that the assignment of the coherent state $|\alpha\rangle$ to a laser field is a good model in most conditions but not in all conditions. The clearest division is that this description is only valid for CW lasers, but there is also a huge cross-section of lasers that are pulsed instead of CW, and those require a multi-mode description to even begin to make sense.
This is part of a wider point, in that lasers are a varied bunch, and there are few properties that hold universally. In the particular case of polarization, most lasers do indeed produce a polarized output, but there are multiple cases that produce unpolarized or partially-polarized light, through a variety of physical mechanisms (like the space-dependent polarization mentioned above, or a time-varying output due to temperature or other fluctuations).
