Yes, it's true, and is easy to prove in this setting. Differentiate under the integral sign to get
$$\frac{d}{dt} \int K(x,y,t)\,dx = \int \partial_t K(x,y,t)\,dx = \int \Delta K(x,y,t)\,dx$$
the Laplacian $\Delta$ taken in the $y$ variable. Now integrating by parts,
$$\int \Delta K(x,y,t)\,dx = \int K(x,y,t) \Delta 1\,dx = 0$$
so that $\int K(x,y,t)\,dx$ is constant with respect to $t$.
To show it's a constant with respect to $y$, you could again differentiate under the integral sign to show that $F(y,t) = \int K(x,y,t)\,dx$ also solves the heat equation, and since it's independent of $t$, it is harmonic. The only continuous harmonic functions on a compact manifold without boundary are constants, by the maximum principle.
More generally, any Riemannian manifold for which this holds is said to be stochastically complete. It's well known (but not quite so easy to prove) that every complete Riemannian manifold with Ricci curvature bounded below is stochastically complete; see for instance
Hsu, Pei, Heat semigroup on a complete Riemannian manifold, Ann. Probab. 17, No. 3, 1248-1254 (1989). ZBL0694.58043.
and references therein.