$\newcommand\R{\mathbb R}$Letting $s:=u-r$ and $t:=v-r$, rewrite the inequality in question as 
\begin{equation*}
	\int_{\R^d}dx\,|p(t,x)-p(s,x)| \le C\Big(\frac ts-1\Big) \tag{0}\label{0}
\end{equation*}
given that $0<s\le t<\infty$. 

Note that 
\begin{equation*}
	|p(t,x)-p(s,x)|\le\int_s^t dw\,|D_w p(w,x)|,
\end{equation*}
where $D_w$ is the operator of partial differentiation with respect to $w$. So, 
\begin{equation*}
		\int_{\R^d}dx\,|p(t,x)-p(s,x)| \le\int_s^t dw\, \int_{\R^d}dx\,|D_w p(w,x)|. \tag{1}\label{1}
\end{equation*}
Next, 
\begin{equation*}
	\int_{\R^d}dx\,D_w p(w,x)=D_w\int_{\R^d}dx\, p(w,x)=D_w1=0
\end{equation*}
and
\begin{equation*}
	D_w p(w,x)=p(w,x)\Big(\frac{|x|^2}{4w^2}-\frac d{2w}\Big). 
\end{equation*}
So, with $z_+:=\max(0,z)$ for real $z$, 
\begin{equation*}
	\int_{\R^d}dx\,|D_w p(w,x)|
	=2\int_{\R^d}dx\,p(w,x)\Big(\frac d{2w}-\frac{|x|^2}{4w^2}\Big)_+
	\le2\int_{\R^d}dx\,p(w,x)\frac d{2w}=\frac dw. 
\end{equation*}

Thus, by \eqref{1}, 
\begin{equation*}
		\int_{\R^d}dx\,|p(t,x)-p(s,x)| \le d\,\ln\frac ts,
\end{equation*}
whence \eqref{0} follows, with $C=d$.