Yes, it could be consistent.
For each of the fundamental forces, we have a certain conserved quantity, which we refer to as a charge. The converse of a result called Noether's theorem tells us that in most cases, a conservation law leads to something called a symmetry$^{\dagger}$, and each symmetry is associated with a mathematical structure called a symmetry group. If we were given a charge, then by studying the symmetry group associated with it, we could learn about the quantum field theoretic interactions that arise from it, and vice versa.
The fundamental forces have the following symmetry groups, respectively:
- Electromagnetism: The very simple unitary group $\text{U}(1)$
- Strong nuclear force: The more complicated special unitary group $\text{SU}(3)$
- Weak nuclear force: The special unitary group $\text{SU}(2)$
We can then learn something about the charges associated with the force and the bosons mediating its interactions. The number of distinct fundamental charges is given by the dimensions of the irreducible representations of the symmetry group, and the number of bosons is given by the number of generators of the group.$^{\ddagger}$ $\text{SU}(2)$ has three generators, and so we have three gauge bosons associated with the weak nuclear force: the $W^+$, $W^-$ and $Z$ bosons. Its representations are two-dimensional, and there are two charges associated with the weak force.
You've simply given us a new conservation law (which at a glance looks "nice" enough for us to be able to apply the converse of Noether's theorem), and therefore a new symmetry group. There's nothing prohibiting us from considering higher-dimensional groups (and as pregunton mentioned, we can go beyond unitary and special unitary groups) that would in turn be associated with new types of charge, giving us new bosons to play around with.
$^{\dagger}$ Noether's theorem itself says that any continuous symmetry has a corresponding conservation law.
$^{\ddagger}$ In particular, the group $\text{SU}(n)$ has $n^2-1$ generators and therefore $n^2-1$ bosons. The group $\text{U}(n)$ has $n^2$ generators and $n^2$ bosons.