The truly minimal condition on $X$ that guarantees that the function $(t,x)\mapsto p_t(x):=\chi_t(x)$ is continuous is tautological: $p_t(x)$ is continuous in $(t,x)$ if and only $p_t(x)$ is continuous in $(t,x)$. As far as the minimality is concerned, I don't think you can do much better than this.
However, one can rather easily see that the sample continuity of $X$ is not enough even for the continuity of $p_t(x)$ in $t$ (for fixed $x$). E.g., let
for real $x$ and real $t\ne0$, with $p_0:=f$, where $f$ is the standard normal pdf. Then $p_t$ is a continuous pdf for each $t$ and, by the Riemann–Lebesgue_lemma,
is continuous in real $t$ for each real $x$. Moreover, $F_0$ is continuous and strictly increasing (in fact, $F_t$ is so for each real $t$). Hence, the process $(X_t)$ defined by the formula
where $U$ is a random variable uniformly distributed on the interval $(0,1)$, has continuous paths. Also, $p_t$ is the pdf of $X_t$, for each $t$. However, $p_t(x)$ is not continuous in $t$ for any real $x\ne0$.