This is a pretty good question. I am curious to see what other commenters say.
My first thought is this. It is true that the EM field fluctuates, and measuring the average value of $\vec{E}(x)$ in some region of space will include statistical randomness. While $\langle \vec{E}(x) \rangle = 0$, there is still uncertainty as $\langle \vec{E}(x)^2 \rangle \neq 0$.
Having said that, one has to be careful about how one pictures a charged particle. If you picture it as a classical point, you may conclude that its position will jitter in a Brownian motion style (which will influence the spread of its wavefunction).
However, in an interacting theory, one must be careful with what one means the definition of an electron. If you want to be as precise as possible, a definite momentum (plane-wave) electron state $| \vec{p} \rangle$ in an interacting theory is an exact energy eigenstate of the theory. In other words, the plane wave simply picks up a uniform phase at it evolves in $t$ and only depends on the energy of the electron. If one makes a wave packet state, with some roughly well localized position, these phases alone will govern how the wavepacket will spread out in time.
In other words, there won't be any special jittering.
I suppose one could imagine that the EM field fluctuations are already taken into account when one defines what exactly one means by an electron in an interacting theory,