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) \tag{1}\label{1} 
\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 $h\in(0,1/2)$. $\quad\Box$ 

---

The OP asked in a [comment][20] if the answer would change if it is additionally assumed that  $X_1=\exp(-U)$ and $Y_1=\exp(-V)$ for random variables $U\ge0$ and $V\ge0$ of which all finite moments exist. 

Then the answer still remains negative, with \eqref{1} replaced by 
\begin{equation*}
	f(x)=\frac{1}{2 x}\,\exp\Big(-\sqrt{\ln\frac1x}\,\Big)\,1(0<x<1),  
\end{equation*}
keeping essentially the same reasoning. 

[10]: https://mathoverflow.net/a/456067/36721 
[20]: https://mathoverflow.net/questions/456103/bounding-x-1-x-1x-2-y-1-y-1y-2-p-by-the-closeness-of-x-and-y/456119#comment1181281_456119