I am aware that there are 6 independent infinitesimal Lorentz transformations that can be separated into 3 rotations and 3 boosts. Is it possible for a quantum field theory to be invariant under the boosts but not invariant under the rotations?
Every quantum field theory has a symmetry group under which its Lagrangian is invariant. Like every group, it must be closed. The boosts are not closed under composition, so they cannot form a symmetry group by themselves.
Any rotation can be achieved by a composition of four boosts, so if each boost leaves the Lagrangian invariant then the resulting rotation will as well.
To add to tparker's answer that the boosts by themselves are not closed under composition, you can either prove this by direct computation - take the product of two boost matrices and you can show that the polar decomposition of the product is generally a boost composed with a nontrivial rotation - or by far the easiest way to prove this is by the Lie correspondence - see Rossmann, "Lie Theory- an Introduction through Linear Groups", section 2.5. Here we learn that the connected analytic subgroups of any connected Lie group correspond bijectively to the Lie subalgebras of the group's Lie algebra. So we only need to witness that the Lie bracket of two infinitesimal boosts is an infinitesimal rotation to prove that the smallest group containing the boosts must needfully include rotations as well.
Indeed, this non closure gives rise to the phenomenon of Thomas Precession / Wigner Rotation.