In a complex, the d orbitals are no longer equal in energy - this is due to ligand field splitting. It will be very difficult to explain this without diagrams, so read a little about crystal field theory/ligand field theory if you are not familiar with this.
Think about B. As I'm sure you agree, the ground electronic configuration of B is [He]2s22p1. Hund's rule does not imply that the lowest energy configuration is [He]2s1 2p2, even though this configuration could have higher multiplicity. Hund's rule can not be used to predict the lowest energy configuration - it says that for a given configuration, the lowest energy term is that with the highest multiplicity - so it says that if the configuration in boron is [He]2s1 2p2 (which it isn't, but bear with me), then the lowest term of that specific configuration is the one in which the two electrons in the p orbitals are unpaired.
Now, Fe(II) has 6 d electrons. But it is not enough to say that the lowest energy configuration is [Ar]3d6, for a complex - because the configuration 3d6 is not specific enough. In a complex, the d orbitals must be considered seperately from each other.
In an octahedral comlex, it turns out that three of the d orbitals are lower in energy (which have t2g symmetry), and the other two d orbitals are higher in energy (which have eg symmetry).
So the label 'd' is useless (if the energy difference is large). We must consider the d orbitals (or "d subshell") as t2g and eg "sub-subshells".
So just as Hund's rule does not predict whether B is [He]2s22p1 or [He]2s1 2p2,
it does not predict whether an octahedral Fe(II) complex is [Ar]t2g6 or [Ar]t2g4 eg2
Sorry, that was a bit long, but hopefully you can see what I'm getting at!