Just adding a function $V(t)$ to the Hamiltonian does nothing - the equations of motion involve only the derivatives of the Hamiltonian w.r.t. $q$ and $p$, and so this changes nothing about the system, you just chose a weirder Hamiltonian for it. Energy is still conserved, it just no longer is the same as the value of the Hamiltonian.
Noether's theorem is not about invariance of the Hamiltonian, it is about invariance of the action, and in the action the addition of a pure function of time to the integrand is an addition of a total time derivative (of the indefinite integral of the added function), which does not change the (in)variance behaviour Noether's theorem cares about.
If you actually want a system in which momentum is conserved but energy is not, you'd need to add a function $V(p,t)$ of momentum and time here, but real world systems do not usually seem to work that way - almost all useful Hamiltonians are of the form $p^2 + V(q,t)$ instead, where $V(q,t)$ is the potential of a possibly time-varying force field.
If you have more than one position $q^i$, then you could also construct a time-variant but momentum-conserving Hamiltonian by adding a function $V(\lvert q^i - q^j\rvert, t)$ to the Hamiltonian. I've never actually seen this done but a toy example might be two devices that become charged over time - the Coulomb force between them would be of this form. Energy is not conserved as there is an influx of charge and hence electric potential, but momentum is conserved, since it's just two bodies attracting/repelling each other with no other forces involved.