I honestly thought about a lot of different possible counterexamples before realizing that a small modification of the structure I already mentioned gives a counterexample. Specifically if we consider the structure of $\mathbb{Z}$ with the successor function and a predicate $P(x,y)$ for $x\equiv y\text{ (mod 2})$. Let $T$ be the theory of this structure.
To see this assume that $T$ has a $\varnothing$-definable strongly minimal imaginary $I$. Since the theory in uncountably categorical there must be a definable finite-to-finite correspondence between $I$ and the definable set $P(x,0)$ (where $0$ is some element of the prime model). Let $\varphi(x,y,\overline{a})$ be that finite-to-finite correspondence (and $\overline{a}$ is a tuple of parameters in the home sort). There needs to be a uniform finite bound $n$ on the sizes of the fibers of the correspondence, so we can write a formula $\psi(w,\overline{z})$ that means '$\varphi(x,y,\overline{z})$ defines a finite-to-finite correspondence between $I$ and $P(x,w)$ with fibers of size at most $n$'. Then the theory $T$ says $\forall w \exists \overline{z} \psi(w,\overline{z})$, so in particular there is some set of parameters $\overline{a}$ in the prime model such that $\psi(x,y,\overline{a})$ is a finite-to-finite correspondence between $I$ and $P(x,0)$. Since all of the prime model is in $\text{dcl}(0)$, we can just let $\overline{a}$ actually be $0$, so we have a formula $\psi(x,y,0)$ that defines a finite-to-finite correspondence between $I$ and $P(x,0)$.
There needs to be a number $m$ such that for any $b\in I$, the largest and smallest element of the home sort related to $b$ by $\varphi(x,y,0)$ are no more than $m$ many successor steps away, so we can actually define a formula $\chi(x,y,0)$ such that each element of $I$ is related to the least element it was related to by $\varphi(x,y,0)$, or $0$ if there were no elements related (this can only happen for finitely many elements of $I$), in other words $\chi(x,y,0)$ defines a function $f(0,x)$ from $I$ into the set $P(x,0)$. The image of this function must be all but finitely many elements of the set $P(x,0)$.
By symmetry, for any even number $2k$, $f(2k,x)$ must define an almost surjective map from $I$ to the set $P(x,0)$. Fix some $2k$. I claim that for all but finitely many $b \in I$, $f(0,x)=f(2k,x)$. If this weren't true, then $T$ would interpret the group $(\mathbb{Z},+)$, which it can't because it's $\omega$-stable. So we can define a new function over the imaginary parameter corresponding to the $P$-equivalence class of $0$, which I'll call $[0]_P$. The function, $g([0]_P,x)$, is given by the average behavior of the function $f(2k,x)$ for different values of $2k$. By stability and symmetry we only need to check some sufficiently large collection of parameters, so the definition of $g([0]_p,x)$ is something like 'there exists more than 17 elements $y$ of $[0]_P$ such that $f(y,x)$ outputs the same value for, let $g([0]_P,x)$ be that value'.
By symmetry again everything we said about the even numbers works for the odd numbers as well and we have a definable function $g([1]_P,x)$ from $I$ into the set $P(x,1)$. Since $[1]_P$ is obtained from $[0]_P$ by shifting everything up by $1$, it must be the case that $g([1]_P,x)=S(g([0]_P,x))$, but by symmetry we also have $g([0]_P,x)=S(g([1]_P,x))$, which is impossible. Therefore no such imaginary can exist.
There's still the matter of the followup question regarding $\varnothing$-definable imaginaries with finite Morley rank, but that's not the title of this question so I'm posting this as an answer rather than as an edit to the question.
