The physical idea is that you'll let $a$ go to infinity for a truly free particle, and if you take this limit, then the specific details of the boundary conditions should be irrelevant, because the boundaries are so far away anyway.
Therefore, you are welcome to choose convenient boundary conditions, and the periodic ones are convenient, because then you have just plain waves $e^{ikx}$, with the admitted $k$-values determined by $e^{ika} = 1$, so $ka = 2\pi n$, and $n \in \mathbb{Z}$.
What you are asking for is $\langle A_\mu(x) \rangle$. Now what could it be? It can't depend on $x$ by translation invariance, so it would have to be a constant four-vector, but as long as the theory is Lorentz invariant there are no constant four-vectors except for $0$.
More concretely, the integral diverges (cubically!) so it must be regularised. Any regularisation which respects Lorentz invariance must give a zero result by the above argument. You might want to try out dimensional regularisation or Pauli-Villars to see this.
EDIT: Expanding a bit. First, you can simplify the propagator by multiplying top and bottom by $\not{k} + m$. The term proportional to $m$ vanishes because of the trace. The trace is then proportional to $k_a$. The integral is (up to a constant factor)
$$\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}\frac{k_{a}}{k^{2}-m^{2}+i\epsilon}$$
This needs to be regularised. Consider Pauli-Villars regularisation. It is easy to show that (since the divergence is cubic) you need two regulator scales to define the integral. We take $\Lambda_1,\Lambda_2\gg m,k$ (and eventually the limit $\Lambda_1,\Lambda_2\rightarrow\infty$), and regulate the integral as
$$\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}} k_a\frac{\left(m^2-\Lambda_1^2\right) \left(m^2-\Lambda_2^2\right)}{\left(k^2-\Lambda_1^2+i \epsilon \right) \left(k^2-\Lambda_2^2+i \epsilon \right) \left(k^2-m^2+i \epsilon \right)}$$
Note that the limit $\Lambda_1,\Lambda_2\rightarrow\infty$ formally gives you back the expression you want and that the whole thing is Lorentz invariant and convergent.
Now it is trivial to show rigourously that the integral vanishes: due to the $k_a$ in the integrand, which is an odd function of $k_a$, while the rest of the integral is even.
This type of argument is often applied without regulating the integral, but the resulting argument would be heuristic, whereas with the regulator it is rigourously correct. You eventually get the hang of when these short-cuts work and when they don't.
Best Answer
As explained by Iwo Bialynicki-Birula in the paper quoted, the Maxwell equations are relativistic equations for a single photon, fully analogous to the Dirac equations for a single electron. By restricting to the positive energy solutions, one gets in both cases an irreducible unitary representation of the full Poincare group, and hence the space of modes of a photon or electron in quantum electrodynamics.
Classical fields are expectation values of quantum fields; but the classically relevant states are the coherent states. Indeed, for a photon, one can associate to each mode a coherent state, and in this state, the expectation value of the e/m field results in the value of the field given by the mode.
For more details, see my lectures
http://arnold-neumaier.at/ms/lightslides.pdf
http://arnold-neumaier.at/ms/optslides.pdf
and Chapter B2: Photons and Electrons of my theoretical physics FAQ.