I) Before we get to quantization and path integrals there are problems already at the classical level. The Legendre transformation is not well-defined without knowledge of the CCR. For instance if the CCRs for the complex bosonic scalar $\hat{\phi}$ and $\hat{\phi}^{\dagger}$ is zero, this would mean that OP's Hamiltonian density ${\cal H}$ is a pure potential term without kinetic terms. The Legendre transformation to the Lagrangian formulation would then become singular.
II) Here is a non-trivial example. Let us instead assume that the CCR for the complex bosonic scalar $\hat{\phi}$ and $\hat{\phi}^{\dagger}$ reads
$$\tag{1} [\hat{\phi}(x,t), \hat{\phi}^{\dagger}(y,t)]
~=~\hbar{\bf 1}~ \delta(x-y), $$
and other CCRs vanish. Equivalently in terms of Poisson brackets
$$\tag{2} \{\phi(x,t), \phi^{\ast}(y,t)\}~=~-i\delta(x-y). $$
We can expand the complex scalar field
$$\tag{3} \phi~=~(\phi^1+i\phi^2)/\sqrt{2} $$
in two real component fields $\phi^a$ , $a=1,2$. Then the CCR (2) becomes
$$\tag{4} \{\phi^1(x,t), \phi^{2}(y,t)\}~=~\delta(x-y). $$
Conclusion: We can identify $\phi^2$ as the momentum for $\phi^1$.
Next recall that OP's Hamiltonian density is (up to a total $x$-derivative)
$$ \tag{5} {\cal H}
~=~\frac{i}{2}\left(\phi\partial_x\phi^{\ast}-\phi^{\ast}\partial_x\phi\right) + V(\phi,\phi^{\ast})
~=~\frac{1}{2}\left(\phi^1\partial_x\phi^2-\phi^2\partial_x\phi^1\right) + V(\phi^1,\phi^2). $$
The corresponding Lagrangian density is then (up to a total $t$-derivative)
$$ \tag{6} {\cal L}~=~\phi^2\dot{\phi}^1-{\cal H}~\sim~
i\phi^*\dot{\phi}-{\cal H}.$$
[Here the $\sim$ symbol means equality modulo total derivative terms.] This Legendre transformation (5)-(6) is explained in detail in this Phys.SE post. Note that the Lagrangian density (6) unconventionally depends on the momentum variable $\phi^2$. Nevertheless, the corresponding action $S=\int \! dt~dx~{\cal L}$ leads to the correct equations of motion, and serves as a starting point for the path integral formulation.
