If $V$ is a *positive* polynomial then $\int_{\mathbb{R}^n} K_t(x,y)dy$ is *strictly less than $1$.* This follows from the Feynman-Kac formula,
$$ \int_{\mathbb{R}^n} K_t(x,y)dy = \mathbb{E}_x \left ( \exp\left ( -\int_0^t V(b(s))ds\right ) \right )$$
where the expectation on the right hand side is over Brownian motion $b$ starting at $x$.
Stochastically you may interpret a positive potential as a local "killing rate" for the process.

On the other hand, since you are concerned with $V$ polynomial and positive, provided $V$ is non-constant, it is standard that $H$ has discrete spectrum and that the ground state eigenvalue $E_0$ is non-degenerate and that the ground state eigenfunction $\psi_0(x)$ vanishes nowhere and may be taken to be positive. Clearly
$$\int_{\mathbb{R}^n} e^{tE_0} K_t(x,y)\psi_0(y)dy = \psi_0(x),$$
which is to say that the heat kernel for $H-E_0$ is stochastically complete on a suitable space.