The Lagrangian $\mathcal{L}$ can be defined as the Legendre transform (when it exists) of the Hamiltonian $\mathcal{H}$, a non-Lorentz scalar quantity (as $\mathcal{H} =T^{00}$). My questions are,
As far as I know, there are no good ways of stating what conditions on the Hamiltonian will cause the Lagrangian to be a Lorentz scalar other than to just say the Hamiltonian must be derived from a Lagrangian that is a Lorentz scalar.
When working with the QFT of the standard model and related QFT's, this is exactly what we do: we start with a Lagrangian, and use Dirac's prescription for canonical quantization to get the Hamiltonian (except when we use path integral methods, in which case we just work directly with the Lagrangian).
As far as relevant cases in High Energy Physics where the Lagrangian is not a Lorentz scalar, Qmechanic correctly points out the case where we are working with a curved background spacetime. Another case is any lattice QFT.
An obvious, kind of dumb, answer is that the Lagrangian corresponding to a given Hamiltonian will be a Lorentz scalar if the Hamiltonian has the form,
$$ \mathcal{H} = \pi^a \frac{\partial}{\partial t} \phi_a(\phi_a,\pi^a) - \mathcal{L}(\phi_a, \partial_i \phi_a,\frac{\partial}{\partial t} \phi_a( \phi_a, \pi^a)) $$
Where $\mathcal{L}(\phi_a, \partial_\mu \phi_a)$ is a Lorentz scalar (here $a$ is just some arbitrary index e.g. spinor, vector, etc.). This is not especially profound as it just follows from the definition of the Legendre transform.
Another example of theories where the lagrangian is not a Lorentz scalar is many of the extensions of QED with magnetic monopoles. These usually contain dependence on an arbitrary direction in space-time, such as the "Dirac String". For complicated reasons, these still lead to Lorentz invariant physical results.
