##\vec{E}## and ##\vec{B}## are vector fields in the sense of the usual rotation group SO(3). Each provides an irreducible representation of this group, because there's not proper subspace that stays invariant under the action of the group.
I'm not sure what you mean bye "decompose the electromagnetic tensor F into the E and B under spatial rotations". Maybe the most simple answer is to use the representation for the full proper orthochronous Lorentz group in terms of ##\mathrm{SO}(3,\mathbb{C})##. This is how the Riemann-Silberstein vector ##\vec{F}=\mathrm{E}+\mathrm{i} \vec{B}## transforms under proper orthochronous Lorentz transformations. The usual rotations are of course represented by the subgroup SO(3) of ##\mathrm{SO}(3,\mathbb{C})##. This means of course that ##\vec{E}## and ##\vec{B}## transform within each other by rotations. A pure rotation-free boost along a given direction is represented by the usual rotation matrix but with an purely imaginary angle, i.e., a boost always mixes electric and magnetic field components.