Consider the follwoing first-order theory:
$$T=\{R(a,b); \forall x,y\,(R(x,y)\rightarrow \exists z\,R(y,z))\}$$
where $a,b$ are constants. Moreover, let us adopt the "standard name assumption", so that we can express every model for $T$ as a set of facts (i.e., ground atoms).
For what I know, $\mathcal{M}=\{R(a,b); R(b,c); R(c,a)\}$ is a minimal model for $T$.
Consider now a model $\mathcal{M}'$ for $T$ for which no such "cycle" on $R$ exists. Then $\mathcal{M}'$ must be infinite. Moreover, no model $\mathcal{M}''$ for $T$ can be such that $\mathcal{M}''\subset\mathcal{M}'$, unless we included "useless" facts in $\mathcal{M}'$ (suppose we didn't).
Then, does it make sense to say that $\mathcal{M}'$ is a minimal model for $T$, despite it is infinite?