The question about quantization of free EM field Let's have the free EM field theory with Coulomb gauge:
$$
\partial^{2}A_{\mu} = 0, \quad A_{0} = 0, \quad (\nabla \cdot \mathbf A ) = 0.
$$
One of the ways of quantizing the field is the following. The expression for energy
$$
\tag 1 W = \frac{1}{2}\int (\mathbf E^{2} + \mathbf B^{2})d^{3}\mathbf r
$$
with a given solution
$$
\mathbf A = \int (\mathbf a (\mathbf k) e^{-ikx}  + \mathbf a^{*} (\mathbf k ) e^{ikx}
) \frac{d^{3}\mathbf k}{\sqrt{(2 \pi )^{3}}}, \quad  \omega_{\mathbf k} = |\mathbf k| 
$$
is rewritten in a form
$$
\tag 2 W = \frac{1}{2}\int \left(\omega^{2}_{\mathbf k}\mathbf Q^{2}(\mathbf k) + \mathbf P^{2}(\mathbf k)\right)d^{3}\mathbf k ,
$$
where
$$
\mathbf Q = \mathbf a + \mathbf a^{*}, \quad \mathbf P = -i\omega_{\mathbf k}(\mathbf a(\mathbf k) - \mathbf a^{*} (\mathbf k )).
$$
So, the question: why do we quantize $(2)$, not $(1)$? I.e., why do we postulate 
$$
[Q_{i}(\mathbf k ) , P_{j} (\mathbf l ) ] = \delta_{ij}\delta (\mathbf k - \mathbf l ),
$$ 
not 
$$[E_{i}(\mathbf x ) , B_{j} (\mathbf x ' ) ] = \delta_{ij}\delta (\mathbf x - \mathbf x' )?
$$
 A: Because the $\textbf{E}$ and $\textbf{B}$ fields are not canonical variables -- it's the $A_\mu$ that appear in the Lagrangian/Hamiltonian. Hence, those are the variables that can have canonical commutator relations. You can only have the commutator relation $[\hat{A}, \hat{B}] = i \hbar$ if $\hat{A}$ and $\hat{B}$ are canonically conjugate.
There's a fairly standard exercise in a lot of textbooks where you figure out the "momentum" for a Lagrangian field theory, and it turns out that it's ill-defined (in fact, it has to do with classical Lagrangian field theories being multi-symplectic, and there's work on this done by Marsden and his crew back in the late '90s that is worth a read if you want a take on it). But, you can still pick one, and call that your conjugate momentum, and because the Lagrangian always involves the vector potential, the momentum has to involve the vector potential as well (actually, its derivatives). The choice to decompose that you list, in terms of plane waves, is most convenient for free fields, but there's nothing stopping you from doing it with spherical waves or some other complete basis, either.
