$\newcommand{\ep}{\varepsilon}$Let
\begin{equation*}
x_n:=\theta_n,\quad s:=r-1>1,\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 $x_0\in[0,1]$. Without loss of generality, $x_0\in(0,1]$.
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*}
Since $c_n/n=x_n\to0$, 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 $h\in(0,1)$. Informally, we are going to show that the sequence $(c_{n-1})$ is mainly confined between the left "moving barrier" $(c^{-h}_n)$ and the right "moving barrier" $(c^h_n)$, where
\begin{equation*}
c^{-h}_n:=\frac n{n-1}\,\frac{1-h}{b_n},\quad c^h_n:=\frac n{n-1}\,\frac{1+h}{b_n}.
\end{equation*}
Moreover, by (2.5) and (2), the jumps of the sequence $(c_{n-1})$ from between these two moving barriers
to the left or right of both of these moving barriers will be of negligible magnitudes.
Indeed, if, for some $n\ge3$, we have
\begin{equation*}
c_{n-1}\le c^{-h}_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 and because $\prod_{j=2}^\infty(1+\frac h{j-1})=\infty$, we will have $c_n\to\infty$ if (3) holds for all natural $k\ge n$, in place of $n$, that is, if $c_{k-1}\le c^{-h}_k$ for all natural $k\ge n$. However, in view of (2), $c^{-h}_k\to\frac{1-h}b<\infty$ as $k\to\infty$. So, (3) cannot hold for all natural $k\ge n$, in place of $n$. So,
there will be some natural $m=m_n\ge n$ such that
\begin{equation*}
c_{n-1}\le c^{-h}_n,\dots,c_{m-1}\le c^{-h}_m,\ c_m>c^{-h}_{m+1}
\end{equation*}
and
\begin{equation*}
c_{n-1}<\cdots<c_{m-1}<c_m.
\end{equation*}
Informally, if $(c_{n-1})$ ventures to the left of the (left) moving barrier $(c^{-h}_n)$, it is returned, in a finite number of steps and in a monotonic manner, to the right of the moving barrier $(c^{-h}_n)$.
On the other hand, similarly, if, for some $n\ge3$, we have
\begin{equation*}
c_{n-1}\ge c^h_n, \tag{3a}
\end{equation*}
then, by (1),
\begin{equation*}
\frac{c_n}{c_{n-1}}\le1-\frac h{n-1}<1. \tag{4a}
\end{equation*}
Therefore, in view of (2) and because $\prod_{j=2}^\infty(1-\frac h{j-1})=0$, there will be some natural $k=k_n\ge n$ such that
\begin{equation*}
c_{n-1}\ge c^h_n,\dots,c_{k-1}\ge c^h_k,\ c_k<c^h_{k+1}
\end{equation*}
and
\begin{equation*}
c_{n-1}>\cdots>c_{k-1}>c_k.
\end{equation*}
Informally, if $(c_{n-1})$ ventures to the right of the (right) moving barrier $(c^h_n)$, it is returned, in a finite number of steps and in a monotonic manner, to the left of the moving barrier $(c^h_n)$.
Recalling now (2.5) and (2), 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.
