$\newcommand{\ep}{\varepsilon}$Let \begin{equation*} x_n:=\theta_n,\quad s:=r-1>2,\quad b:=\frac{s-1}{2s}>0, \end{equation*} \begin{equation*} f(x):=1-(1-x/s)^s, \end{equation*} so that \begin{equation*} x_n=f(x_{n-1}) \end{equation*} for natural $n$, with $x_0\in[0,1]$. We have $f'(x)=(1-x/s)^{s-1}<1$ for $x\in(0,1]$ and hence $f(x)<x$ for $x\in(0,1]$, so that $x_n$ is decreasing to some limit $x_\infty=f(x_\infty)\in[0,1]$. Therefore and because $f(x)<x$ for $x\in(0,1]$, we have $x_\infty=0$, so that $x_n\downarrow0$ (as $n\to\infty$).
Next, $f(x)=x-(b+o(1))x^2$ as $x\downarrow0$. So, \begin{equation*} x_n=x_{n-1}-a_n x_{n-1}^2 \end{equation*} for some $a_n\to b$ and all natural $n$. Letting now \begin{equation*} c_n:=nx_n, \end{equation*} we have \begin{equation*} c_n=\frac n{n-1}\,c_{n-1}-\frac{b_n}n\,c_{n-1}^2 \tag{1} \end{equation*} for some \begin{equation*} b_n\to b. \tag{2} \end{equation*}
Take any real $\ep>0$. If $c_{n-1}\ge(n-1)\ep$ for some natural $n$, then, by (1),
\begin{equation*}
\frac{c_n}{c_{n-1}}\le\frac n{n-1}-\frac{b_n}n\,(n-1)\ep\to1-b\ep
\end{equation*}
(as $n\to\infty$). Therefore and in view of the (following from (1)) "upper semi-continuity" condition $c_n\le\frac n{n-1}\,c_{n-1}$,
we have $c_n/n\to0$. So, by (1) and (2), we get the crucial conclusion that
\begin{equation*}
\frac{c_n}{c_{n-1}}\to1. \tag{2.5}
\end{equation*}
Take now any $h\in(0,1)$.
If, for some $n\ge3$, we have
\begin{equation*}
c_{n-1}\le\frac n{n-1}\,\frac{1-h}{b_n}, \tag{3}
\end{equation*}
then, by (1),
\begin{equation*}
\frac{c_n}{c_{n-1}}\ge1+\frac h{n-1}>1. \tag{4}
\end{equation*}
Therefore, in view of (2) and because $\prod_{j=2}^\infty(1+\frac h{j-1})=\infty$, there will be some smallest natural $m=m_n\ge n$ such that
\begin{equation*}
c_m>\frac{m+1}m\,\frac{1-h}{b_{m+1}}.
\end{equation*}
On the other hand, (3) holds with $m$ in place of $n$. Thus, in view of (2.5) and (2),
\begin{equation*}
c_m=c_{m_n}\to\frac{1-h}b
\end{equation*}
along any increasing sequence of values of $n$ satisfying (3).
On the other hand, similarly, if, for some $n\ge3$, we have
\begin{equation*}
c_{n-1}\ge\frac n{n-1}\,\frac{1+h}{b_n}, \tag{3a}
\end{equation*}
then, by (1),
\begin{equation*}
\frac{c_n}{c_{n-1}}\le1-\frac h{n-1}<1. \tag{4}
\end{equation*}
Therefore, in view of (2) and because $\prod_{j=2}^\infty(1-\frac h{j-1})=0$, there will be some smallest natural $k=k_n\ge n$ such that
\begin{equation*}
c_k<\frac{k+1}k\,\frac{1+h}{b_{k+1}}.
\end{equation*}
On the other hand, (3a) holds with $k$ in place of $n$. Thus, in view of (2.5) and (2),
\begin{equation*}
c_k=c_{k_n}\to\frac{1+h}b
\end{equation*}
along any increasing sequence of values of $n$ satisfying (3a).
We conclude that, for any $h\in(0,1)$, \begin{equation} \frac{1-h}b\le\liminf_n c_n\le\limsup_n c_n\le\frac{1+h}b; \end{equation} that is, $c_n\to1/b$; that is, $\theta_n=x_n\sim1/(nb)$, as desired.