Such a bound does not exist. 

This is shown by the following modification of the [previous example][10]: 

Suppose that $X_1,X_2$ are independent random variables each with the pdf given by the formula 
\begin{equation}
	f(x)=\frac c{x\ln^2 x}\,1(0<x<1/2) 
\end{equation}
for $c:=1/\int_0^{1/2}\frac{dx}{x\ln^2 x}$; 
$Y_1=h+(1-h)X_1$ and $Y_2=h+(1-h)X_2$, where $h\in(0,1/2)$. 

Then $0\le Y_1-X_1\le h$ and hence 
\begin{equation}
	\|X_1-Y_1\|_r\le h  
\end{equation}
for all real $r>0$. 

On the other hand, writing $A\gg B$ if $B=O(A)$, we get 
\begin{equation}
	\Big|\frac{X_1}{X_1+X_2}-\frac{Y_1}{Y_1+Y_2}\Big|
	=\frac{h|X_1-X_2|}{(X_1+X_2)(2h+(1-h)(X_1+X_2))}
	\gg1
\end{equation}
on the event $\{X_2\le X_1/2\le h/2\}$. So, 
\begin{equation}
	\Big\|\frac{X_1}{X_1+X_2}-\frac{Y_1}{Y_1+Y_2}\Big\|_p^p
	\gg\int_0^h dx_1\,f(x_1)
	\int_0^{x_1/2} dx_2\,f(x_2)\asymp\frac1{\ln^2 h}. 
\end{equation}
So, letting $h\downarrow0$, we see that the inequality 
\begin{equation}
	\Big\|\frac{X_1}{X_1+X_2}-\frac{Y_1}{Y_1+Y_2}\Big\|_p\le C\|X_1-Y_1\|_r^\alpha
\end{equation}
cannot hold for any real $C$ (even if $C$ depends on the distribution of $X_1$), any real $p>0$, any real $r>0$, any real $\alpha>0$, and all $h\in(0,1/2)$. $\quad\Box$ 

[10]: https://mathoverflow.net/a/456067/36721