Do fields describing different particles always commute? Is it true that field operators describing different particles (for example a scalar field operator $\phi (x) $ and a spinor field operator $\psi (x) $) always commute (i.e. $ [\phi (x), \psi (y) ]=0, \forall x,y $) in interacting theory?
Or is it true only at equal times? (i.e. $ [\phi (t,\vec x), \psi (t, \vec y) ]=0, \forall \vec x, \vec y $)
Or is it in general not true even at equal times?
Finally, if the fields in account are both fermionic must the commutator be replaced with an anticommutator?
 A: lurscher asks for a physical interpretation of CR Drost's correct answer. I'm answering separately because my response was too long to fit into a comment.
In the trivial case where the two fields are completely uncoupled - either directly or indirectly by way of both being coupled to some third field - the Heisenberg annihilators commute at all times. Otherwise, they have nonzero commutators at different times. We can interpret this second case physically by noting that virtual-particle loops of the second field's particle appear in the exact propagator for the first field. Roughly speaking, if you create a particle $A$ that can scatter/decay into virtual particles of type $B$ by some later time, then in creating particle $A$ there is some amplitude to indirectly create a particle $B$ as well, so the two particles' creation operators are not independent.
A: The Schrödinger-picture annihilators for different sorts of particle commute; so e.g. if you've got creation and annihilation operators for neutrinos as well as electrons in some sort of box, those operators commute across species but within one species do not commute with their adjoints.
Note that if you add time-dependence to them to put them into the Heisenberg picture, this is no longer true in general. Several interaction pictures do have this property, in particular when the baseline Hamiltonian that we "Heisenbergize" for lack of a better term has the quadratic form $\hbar\omega~\hat a^\dagger \hat a$ so that $\hat a(t)$ takes on the form $e^{-i\omega t}~\hat a_0.$
A: No.
In full generality, the super-commutator of two fundamental fields is identical to the Dirac bracket of the corresponding classical variables (modulo the standard obstructions). If the system is unconstrained, then the Dirac bracket agrees with the Poisson bracket, which means that two independent fields super-commute. But if there are non-trivial constraints, then the Dirac bracket may very well mix different fields, so that the corresponding operators will fail to super-commute.
As a trivial example, consider the non-zero commutator of the scalar potential and the Dirac field in QED in the Coulomb gauge (see eq. 15.11 in Bjorken&Drell), or how a Stückelberg field fails to commute with its associated B-field (see eq.38 in 1510.03213). More generally, the Nakanishi-Laudrup field does not usually commute with the rest of fields of the theory.
