$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\De}{\Delta}\newcommand\R{\mathbb R}$Edit: This answer is insufficient, even though (almost) all the reasoning appears relevant to the problem. I will try to come back and fix it.
Suppose that $T\in(0,1]$.
Integrating in $y$, simplify the expression for $F$ as follows:
\begin{equation*}
F[h](s)=I_1(h)+I_2(h), \tag{1}
\end{equation*}
where
\begin{equation*}
I_1(h):=\int_0^s \frac{1}{2} \Big(\text{erf}\Big(\frac{s-u}{2 \sqrt{A_h(s)-A_h(u)}}\Big)+1\Big) h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A_h(u)\big)^2}\, du,
\end{equation*}
\begin{equation*}
I_2(h):=\int_0^{\infty} \frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A_h(s)}}\Big)+1\Big)\rho(x)\,dx.
\end{equation*}
Here the notation $A_h$ reminds us that $A$ depends on $h$.
Let $H_T:=\mathcal H_T$. Let $B_T$ be the set of all bounded real-valued functions on $[0,T]$, endowed with the sup-norm.
Since $0\le h\le1$, the operator that maps any $h\in H_T$ to the function $[0,T]\ni u\mapsto\big(1+h(u)\big)^2$ in $L_T$ is Lipschitz with a universal Lipschitz constant.
Take any $h_0,h_1$ in $H_T$ with
\begin{equation*}
\de:=\|h_1-h_0\|<1/16. \tag{2}
\end{equation*}
Letting
\begin{equation*}
J_h(t):=\int_0^t (1+h(r))^2\,dr,
\end{equation*}
we see that $J_h$ is $4$-Lipschitz, $t\le J_h(t)\le4t$, $A_h$ is $1$-Lipschitz, $|J_{h_1}(t) -J_{h_0}(t)|\le8t\de$, and hence
\begin{equation*}
\frac u4\le A_{h_j}(u)\le\frac{A_{h_{1-j}(u)}}{1-8\de}\le\frac u{1-8\de}, \tag{3}
\end{equation*}
\begin{equation*}
|A_{h_1}(u)-A_{h_0}(u)|\le\frac{8\de u}{1-8\de}\le16\de u; \tag{4}
\end{equation*}
here in what follows, $0\le t\le T$, $0<u<s\le T$, and $j\in\{0,1\}$, unless specified otherwise.
For brevity, let us refer to the Lipschitz maps with Lipschitz constants depending only on $\de$ as good-Lipschitz. In particular, the map $H_T\ni h\mapsto A_h$ is good-Lipschitz. Therefore, the maps $H_T\ni h\mapsto\big(1+h\circ A_h(u)\big)^2$ and
\begin{equation*}
H_T\ni h\mapsto h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A_h(u)\big)^2} \tag{5}
\end{equation*}
are good-Lipschitz, for each $u\in[0,T]$. The latter map is also bounded by a constant depending only on $\de$; let us refer to such maps as well-bounded.
For $h_t:=h_0+t(h_1-h_0)$, $\tau_t:=A_{h_t}(s)-A_{h_t}(u)$, and $\De\tau:=\tau_1-\tau_0$, we have
$0<\frac{s-u}4\le\tau_t\le s-u$ (since $J_h$ is $4$-Lipschitz and $A_h$ is $1$-Lipschitz), $|\De\tau|\le32\de s$ (by (4)), and hence
the absolute value of the derivative of the map
\begin{equation*}
[0,1]\ni t\mapsto
\dfrac{1}{2} \Big(\text{erf}\Big(\dfrac{s-u}{2 \sqrt{A_{h_t}(s)-A_{h_t}(u)}}\Big)+1\Big)
\end{equation*}
is
\begin{equation*}
\begin{aligned}
& \frac{ s-u }{4 \sqrt{\pi } \tau_t^{3/2}}\ \exp\Big\{-\frac{(s-u)^2}{4\tau_t}\Big\}\,|\De\tau| \\
& \le\frac1{4^{-1/2} \sqrt{\pi } (s-u)^{1/2}}\ \exp\Big\{-\frac{s-u}4\Big\}\,32\de s.
\end{aligned}
\end{equation*}
The integral in $u\in(0,s)$ of the latter expression is $\le CT$; here and in what follows, $C$ will denote various well-bounded expressions.
Therefore and because $|\text{erf}|\le1$ and the map (5) is good-Lipschitz and well-bounded, we conclude that $I_1$ is $CT$-Lipschitz.
It is similar but easier to show that $I_2$ is $CT$-Lipschitz as well.
Indeed, still with $h_t:=h_0+t(h_1-h_0)$, let $\rho_t:=A_{h_t}(s)$, and $\De\rho:=\rho_1-\rho_0$.
Then
$0<\frac s4\le\rho_t\le s$ (since $J_h$ is $4$-Lipschitz and $A_h$ is $1$-Lipschitz), $|\De\rho|\le16\de s$ (by (4)), and hence
the absolute value of the derivative of the map
\begin{equation*}
[0,1]\ni t\mapsto
\frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A_{h_t}(s)}}\Big)+1\Big)
\end{equation*}
is
\begin{equation*}
\begin{aligned}
&\frac{|s+x|}{4 \sqrt{\pi } \rho_t^{3/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho_t}\Big\}\,|\De\rho| \\
&=\frac{|s+x|}{4 \sqrt{\pi } \rho_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho_t}\Big\}\,
\frac{|\De\rho|}{\rho_t} \\
&\le\frac{|s+x|}{4 \sqrt{\pi } \rho_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho_t}\Big\}\,
64\de\le C;
\end{aligned}
\end{equation*}
so, the integral in $u\in(0,s)$ of the latter expression is $\le CT$. So, $I_2$ is $CT$-Lipschitz as well.
We conclude that $F$ is $CT$-Lipschitz and hence a contraction if $T$ is small enough.
Best Answer
$\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$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 $T>0$.