$\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$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 $\la\in(0,1]$. $\quad\Box$