Is Br sp³-hybridised in HBr?
From a theoretical viewpoint we would expect the Br to be sp3 hybridized, because in the HBr molecule the Br atom has 4 pairs of electrons around it (the 3 lone pairs plus the bonding pair), and symmetry (plus the repulsion of electrons for each other) tells us to expect that 4 pairs of electrons arrange themselves in a tetrahedron, and as you know we explain a tetrahedral arrangement of electron pairs around an atom by arguing the s and p atomic orbitals form sp3 hybrid orbitals.

It's the same valence shell electron pair repulsion (VSEPR) logic that leads us  to expect the 4 bonding pairs around the C in CH4 should be arranged tetrahedrally, and, therefore, the C atom is sp3 hybridized.  (We would also expect that a more careful and precise determination would give us some deviation from this prediction for HBr, since unlike CH4 the 4 electron pairs around the Br in HBr are not equivalent.)

But you are interested in the question of whether there could be some experimental verification of this expectation.  I don't think the experimental dipole moment would work, because it's too crude a picture of the electron distribution, it just tells us whether the electrons spend extra time near the H or near the Br, and we already know it will be the Br because of its much stronger effective nuclear charge for valence electrons.  What you want to know is something like whether the valence electrons spend significantly extra time on the "far" side of the Br, away from the H-Br bond, as opposed to the "near" side, so we can conclude they're in sp3 hybrid orbitals that point away from the H-Br bond.  Offhand I can't think of an experiment that would unambiguously test this.

You could certainly calculate the electron density in the HBr molecule, with a good quantum chemistry package, and for a molecule this small the results should be quite reliable.  That would tell you whether it makes more sense to consider the Br lone pairs as in sp3 hybrid orbitals or atomic p orbitals.