The problem with the thought experiment is, as the comments already stated, that the propagator is not a physical quantity per se.
The (Feynman) propagator is defined as the time-ordered VEV of two fields (from Wikipedia)
$$ D_F(x-y) = i \langle 0 | T \, \Phi(x) \Phi(y) | 0 \rangle $$
The fact that this does not vanish over spacelike intervals is one of the puzzling features of QFT. One must not forget, however, that bra and ket in the above equation are the vacuum state. What the propagator describes (in this fashion) is a vacuum fluctuation.
In order to make contact to real particles, one has to invoke the LSZ reduction mashinery. What you actually want is for $\langle x | y \rangle$ to exist over spacelike separations, where $| x \rangle$ is the position eigenstate at Bob's oven and $| y \rangle$ is the position eigenstate at Alice's detector. In order to evalute this bracket, the LSZ formula will inevitably give you equation-of-motion operators (like $\square + m^2$) acting on the field operators that will, in the end, reside between vacuum states, which gives the link to the propagator.
$$\langle x | y \rangle = (\square_x + m^2) (\square_y + m^2) \langle 0 | T\, \Phi(x) \Phi(y) | 0 \rangle$$
(Note: The LSZ formula is usually given with momentum eigenstates in mind, I'm not sure whether I adopted it correctly to the position eigenstate case)
These operators, acting on the propagator, which might not vanish for $x-y$ spacelike, ensure that the actual matrix element DOES vanish for all spacelike intervals. And since it is the matrix elements that carry the physical information, QFT does not violate special relativity on which it is based.