$\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$The answer is no.
E.g., suppose that for all small enough $t>0$ we have
\begin{equation*}
f(t)=t,\quad g(t)=0,\quad \al'(t)=-\frac1{\sqrt t}\,\Big(1+\sin\frac1t\Big),\quad \be'(t)=-\frac1{\sqrt t},
\end{equation*}
with $\al(0)=1=\be(0)$.
Then all your conditions hold. In particular, for $t\downarrow0$,
\begin{equation*}
|\al(t)-\be(t)|=\Big|\int_{1/t}^\infty\frac{\sin z}{z^{3/2}}\,dz\Big|=O(t^{3/2})
\end{equation*}
and
\begin{equation*}
\int_0^t|\al'(s)-\be'(s)|\,ds=\int_{1/t}^\infty\frac{|\sin z|}{z^{3/2}}\,dz\asymp t^{1/2},
\end{equation*}
so that
\begin{equation*}
\begin{aligned}
\|\al-\be\|_t &\le C\sqrt{t} \big(\|\al-\be\|_t+\|f-g\|_t\big) \\
\|f-g\|_t &\le C\sqrt{t}\left(\|\al-\be\|_t+\|f-g\|_t + \int_0^t|\al'(s)-\be'(s)|ds\right)
\end{aligned}
\tag{1}\label{1}
\end{equation*}
for some real $C>0$ and all $t$ in some right neighborhood of $0$. So, inequalities \eqref{1} hold for some real $C>0$ and all real $t\ge0$.
Yet, $f\ne g$ in any right neighborhood of $0$. So, $(\al,f)\ne(\be,g)$ on $[0,T]$, for any real $T>0$.
$\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$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, $U=1$, $v=1$, and $u=0$, 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 $\la\in(0,1]$. $\quad\Box$
Best Answer
$\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand\ep\epsilon$The integral in question is \begin{equation*} I(u,\al):=\int_0^u \frac{dr}r \int_{[-1,1]^2} dy \int_{[-1,1]^2} dx\,e^{-\al^2|x-y|^2/r}. \end{equation*} Using the substitution $r=\al^2 s$ (suggested by Giorgio Metafune), we get \begin{equation*} I(u,\al)=J(u/\al^2), \end{equation*} where \begin{equation*} J(v):=\int_0^v \frac{ds}s \int_{[-1,1]^2} dy \int_{[-1,1]^2} dx\,e^{-|x-y|^2/s} =\int_0^v \frac{ds}s\, K(s)^2, \end{equation*} where \begin{equation*} K(s):=\int_{-1}^1 da\int_{-1}^1 db\,e^{-(a-b)^2/s} \le\int_{-1}^1 da\int_{-\infty}^\infty db\,e^{-(a-b)^2/s} =\int_{-1}^1 da\, \frac c2\,\sqrt s=c\sqrt s; \end{equation*} here and in what follows $c$ denotes universal positive real constants (possibly different even within the same formula), and $s$ is any positive real number. Also, \begin{equation*} K(s)\le\int_{-1}^1 da\int_{-1}^1 db=4. \end{equation*} So, \begin{equation*} K(s)^2\le c\min(1,s). \end{equation*} So, \begin{equation*} J(v)\le c\int_0^v \frac{ds}s\,\min(1,s)=c(\min(1,v)+\ln\max(1,v))\le c\ln(1+ v); \end{equation*} the latter inequality follows because (say) $\min(1,v)+\ln\max(1,v)$ and $\ln(1+ v)$ are (i) positive and continuous in $v\in(0,\infty)$ and (ii) asymptotically equivalent to each other as $v\downarrow0$ and as $v\to\infty$.
So, \begin{equation*} I(u,\al)\le c\ln\Big(1+\frac u{\al^2}\Big). \end{equation*}
So, upon the substitution $u=\al^2 v$, the inequality in question reduces to the inequality
\begin{equation*} \ln(1+v)\le C v^\ep\al^{2\ep-2\be} \tag{1}\label{1} \end{equation*} for some real $C>0$, some real $\ep>0$, all real $v>0$, and all $\al\in(0,1]$. If we now take $\ep=\min(1,\be)$, then \eqref{1} will hold for some real $C>0$ depending only on $\be$. $\quad\Box$