$\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$What you wanted us to prove is not true. 

Indeed, take any $\phi\in\EE$ such that $\phi\ge1_{[-1/2,1/2]}$. Write $A\gg B$ for $A\ge cB$, where $c$ is a universal positive real constant. 

Then, for $w:=x-y_2$,  
\begin{equation}
\begin{aligned}
	&\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1 \\ 
	&\gg\frac1\la\,\int_\R dy_1\, 1(|y_1-x|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(y_1-y_2)^2}{4r} \\ 	
	&=\frac1\la\,\int_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(w+z)^2}{4r} \\ 	&\ge\frac1\la\,\int_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{w^2+z^2}{2r} \\ 
	&\ge\exp\Big(-\frac{\la^2}{8r}\Big)\frac1{\sqrt r}\,\exp-\frac{w^2}{2r}.  \end{aligned}
\end{equation}	
So, 
\begin{equation}
\begin{aligned}
	&\int_\R dy_2\,\Big(\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1\Big)^2 \\ 
	&\gg \exp\Big(-\frac{\la^2}{4r}\Big) 
	\int_\R dw\,\frac1r\,\exp-\frac{w^2}r \\ 
	&\gg \frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big)
\end{aligned}
\end{equation}
and hence 
\begin{equation}
\begin{aligned}
	I&:=\int_0^{|v-u|}dr\,\int_\R dy_2\,\Big(\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1\Big)^2 \\ 
	&\gg \int_0^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\ 
	&\ge \int_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\ 
	&\ge \int_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{2|v-u|}\Big) \\ 
	&\gg |v-u|^{1/2}\exp\Big(-\frac{\la^2}{2|v-u|}\Big). 
\end{aligned}
\end{equation}

Letting now, for instance, $U=1$, $u=1$, and $v=0$, for all $\la\in(0,1]$ we get 
\begin{equation}
	I\gg1. 
\end{equation}
So, if $\beta<1/2$, then there is no real $\ep>0$ and $C>0$ such that $I^{1/2}\le C|v-u|^{\ep} \la^{1/2-\beta}$ for all $\la\in(0,1]$. $\quad\Box$