Such a bound does not exist.

This is shown by the following modification of the previous example:

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 $h\in(0,1/2)$. $\quad\Box$

Such a bound does not exist.

This is shown by the following modification of the previous example:

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 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.