Your motivation is about aperiodicity implying high complexity, but you ask whether aperiodicity implies an upper bound on complexity, which is a little strange. Probably there is a typo, and $O(g(r))$ should be $\Omega(g(r))$?

In any case, I'll give a simple generalization of the Morse-Hedlund theorem to all groups.

Definition. Let $G$ be any group, and let $D_1, D_2, D_3, ...$ be a family of finite sets, such that $|D_i| \geq 1$. We say $(D_i)_i$ satisfies property (C) (for "connected") if the following holds: For $i \geq 1$, define a graph $\mathcal{G}_i$ with nodes $G$, and an edge between $g$ and $h$ if and only if $gD_i \cup hD_i \subset kD_{i+1}$ for some $k \in G$. Then for all $i$, the graph $\mathcal{G}_i$ is connected.

For $G = \mathbb{Z}$, we can take $D_i = [1,i]$. For general groups we can take balls of increasing radius $r$ with respect to any finite symmetric generating set.

Let $X \subset A^G$ be any subshift (closed set invariant under $G$-translations). Then we can measure complexity of $X$ by counting the finite sets $P_n = \{x|_{D_i} \;|\; x \in X\}$.

Theorem. If $X$ is an infinite subshift, $(D_i)_i$ have property (C), and $P_i$ are the associated pattern sets, then $|P_n| \geq n+1$ for all $n$.

Proof. Clearly $|P_1| \geq 2$. Otherwise, since $|D_1| \geq 1$, in particular all configurations in $X$ have the same symbol at the origin, thus $|X| \leq 1$ and $X$ is not infinite.

We now show $|P_{i+1}| > |P_i|$, from which $|P_n| \geq n+1$ immediately follows by induction.

Suppose the contrary, that $|P_{i+1}| \leq |P_i|$ for some $i$. Since $D_i \subset D_{i+1}$, we actually have $|P_{i+1}| = |P_i|$, and we see that by shift-invariance that for any $x \in X$, the restriction $x|_{gD_i}$ uniquely determines $x|_{hD_{i+1}}$ whenever $gD_i \subset hD_{i+1}$.

I claim that then $x|_{D_i}$ uniquely determines $x$ for all $x \in X$. This is more or less immediate from (C): Write $N \subset \mathcal{G}_i$ for all $g$ such that $x|_{D_i}$ determines $x|_{gD_i}$ uniquely. Note that ``$x|_{aD_i}$ determines $x|_{bD_i}$ uniquely'' is a transitive relation.

By definition $N$ contains $1_G$. It is also a connected component of $\mathcal{G}$: if we have the edge $(g, h)$ then $gD_i \cup hD_i \subset kD_{i+1}$ for some $k \in G$, thus $x|_{gD_i}$ determines $x|_{kD_{i+1}}$, in particular $x|_{gD_i}$ determines $x|_{hD_i}$. Square.

In the case of $\mathbb{Z}$, with the choice $D_i = [1, i]$ this precisely recovers the Morse-Hedlund theorem. In the case of $\mathbb{Z}^2$ if we use rectangles for $D_i$ that grow alternately in the two dimensions, then this gives only a linear lower bound on complexity of squares, so it falls far behind what is known about Nivat's conjecture.

Your motivation is about aperiodicity implying high complexity, but you ask whether aperiodicity implies an upper bound on complexity, which is a little strange. Probably there is a typo, and $O(g(r))$ should be $\Omega(g(r))$?

In any case, I'll give a simple generalization of the Morse-Hedlund theorem to all groups.

Definition. Let $G$ be any group, and let $D_1, D_2, D_3, ...$ be a family of finite subsets of $G$, such that $|D_i| \geq 1$. We say $(D_i)_i$ satisfies property (C) (for "connected") if the following holds: For $i \geq 1$, define a graph $\mathcal{G}_i$ with nodes $G$, and an edge between $g$ and $h$ if and only if $gD_i \cup hD_i \subset kD_{i+1}$ for some $k \in G$. Then for all $i$, the graph $\mathcal{G}_i$ is connected.

For $G = \mathbb{Z}$, we can take $D_i = [1,i]$. For general groups we can take balls of increasing radius $r$ with respect to any finite symmetric generating set.

Let $X \subset A^G$ be any subshift (closed set invariant under $G$-translations). Then we can measure complexity of $X$ by counting the finite sets $P_i = \{x|_{D_i} \;|\; x \in X\}$.

Theorem. If $X$ is an infinite subshift, $(D_i)_i$ have property (C), and $P_i$ are the associated pattern sets, then $|P_n| \geq n+1$ for all $n$.

Proof. Clearly $|P_1| \geq 2$. Otherwise, since $|D_1| \geq 1$, in particular all configurations in $X$ have the same symbol at the origin, thus $|X| \leq 1$ and $X$ is not infinite.

We now show $|P_{i+1}| > |P_i|$, from which $|P_n| \geq n+1$ immediately follows by induction.

Suppose the contrary, that $|P_{i+1}| \leq |P_i|$ for some $i$. Since $D_i \subset D_{i+1}$, we actually have $|P_{i+1}| = |P_i|$, and we see that by shift-invariance that for any $x \in X$, the restriction $x|_{gD_i}$ uniquely determines $x|_{hD_{i+1}}$ whenever $gD_i \subset hD_{i+1}$.

I claim that then $x|_{D_i}$ uniquely determines $x$ for all $x \in X$. This is more or less immediate from (C): Write $N \subset \mathcal{G}_i$ for all $g$ such that $x|_{D_i}$ determines $x|_{gD_i}$ uniquely. Note that ``$x|_{aD_i}$ determines $x|_{bD_i}$ uniquely'' is a transitive relation.

By definition $N$ contains $1_G$. It is also a connected component of $\mathcal{G}$: if we have the edge $(g, h)$ then $gD_i \cup hD_i \subset kD_{i+1}$ for some $k \in G$, thus $x|_{gD_i}$ determines $x|_{kD_{i+1}}$, in particular $x|_{gD_i}$ determines $x|_{hD_i}$. Square.

In the case of $\mathbb{Z}$, with the choice $D_i = [1, i]$ this precisely recovers the Morse-Hedlund theorem. In the case of $\mathbb{Z}^2$ if we use rectangles for $D_i$ that grow alternately in the two dimensions, then this gives only a linear lower bound on complexity of squares, so it falls far behind what is known about Nivat's conjecture.