Your statement of the Pitman-Koopman-Darmois theorem is off; there is an additional assumption that the support of $\mathcal X$ does not change as $\theta$ changes where $\mathcal X$ is the support of $X_1$ and $\theta$ parameterizes the family. A quick counterexample to the statement of the theorem as given in the OP is the family $\{\mbox{Uniform}(0, \theta): \theta > 0\}$ for which $\max\{X_j, 1 \le j \le n\}$ is sufficient and does not vary with the sample size $n$.
More in the spirit of your question, the answer is yes, there do exist distributions who sufficient statistics are "lossy" even when the conditions of the PKD theorem are satisfied. Consider $X_1, X_2, ...$ iid from a Gamma distribution with shape parameter $\alpha$ (known) and mean $\mu$, and $Z_1, Z_2, ...$ iid Bernoulli with success probability $p$ also known. Then take $Y_i = Z_i X_i - (1 - Z_i)$, and our sample becomes $Y_1, Y_2, ...$. We only get information about $\mu$ when $Y_i \ne -1$, so our sufficient statistic is $\sum_{i: Y_i \ne -1} Y_i$ which grows like $pn$ on average.
Sublinear growth is possible, proceeding along the train of thought suggested above, i.e. using mixtures of distribution, and indeed this is getting at something that is useful in practice. Take $(X_1, Z_1), (X_2, Z_2), ...$ to be iid distributed according to an infinite mixture of normals $f(x, z | \pi, \mu) = \prod_{i = k} ^ \infty [\pi_k N(x | \mu_k, 1)]^{I(Z_i = k)}$, with $Z_i$ an indicator of which cluster $X_i$ is in (I'm not sure if the representation of it via a density that I wrote is valid but you should get the general idea); the dimension of the sufficient statistics should increase only when new clusters are discovered, and the rate of the appearance of new clusters can be controlled by taking $\{\pi_k\}_{k = 1} ^ \infty$ to be known and carefully choosing them; my hunch is that it should be easy to make it grow at a rate of $\log(1 + Cn)$ since I think this is how fast the number of clusters grows in the Dirichlet process.
Refer to the lecture notes here on page 5.
Joint density of the sample $ X=(X_1,X_2,\ldots,X_n)$ for $\theta\in\mathbb R$ is as you say $$f_{\theta}( x)=\mathbf1_{\theta<x_{(1)},x_{(n)}<\theta+1}=\mathbf1_{x_{(n)}-1<\theta<x_{(1)}}\quad,\,x=(x_1,\ldots,x_n)$$
where $x_{(1)}=\min_{1\le i\le n}x_i$ and $x_{(n)}=\max_{\le i\le n}x_i$.
It is clear that $T(x)=(x_{(1)},x_{(n)})$ is sufficient for $\theta$ by the Factorization theorem.
Define $A_x=(x_{(n)}-1,x_{(1)})$ and $A_y=(y_{(n)}-1,y_{(1)})$.
Then for some $y=(y_1,\ldots,y_n)$, observe that the ratio $f_{\theta}(x)/f_{\theta}(y)$ takes the simple form
$$\frac{f_{\theta}(x)}{f_{\theta}(y)}=\frac{\mathbf1_{\theta\in A_x}}{\mathbf1_{\theta\in A_y}}=\begin{cases}0&,\text{ if }\theta\notin A_x,\theta\in A_y \\ 1&,\text{ if }\theta\in A_x,\theta\in A_y \\ \infty &,\text{ if }\theta\in A_x,\theta\notin A_y\end{cases}$$
Clearly this is independent of $\theta$ if and only if $A_x=A_y$, that is iff $T(x)=T(y)$, which proves $T$ is indeed minimal sufficient.
Another proof using the definition of minimal sufficiency is given on page 3 of the linked notes.
As this example shows, there is no such rule of thumb in general for ascertaining minimal sufficiency of a statistic simply by comparing the dimensions of the statistic and that of the parameter.
Best Answer
Are you given that $r$ is known? If it is, then you could define $h(x)$ as you have but including the product in the numerator. This would not be a problem as $r$ is known.