Is the unit group of any finitely generated reduced $\Bbb Z$ algebra finitely generated? If $A$ is  finitely generated commutative reduced $\Bbb Z$ algebra, must the unit group $A^{\times}$ be finitely generated?
The question is motivated by the Dirichlet unit theorem which says the unit group of the algebraic integer ring of any number field is finitely generated. And for other $\Bbb Z$ algebras such as finite fields, their unit groups are even finite. 

"I think the case that $A$ is finite as $\mathbf Z$-module is always true".
Yes, it's true, it's even presented as "a generalization of the unit theorem" in §4.7 of P. Samuel's booklet on ANT. The particular case that $A$ is an integer domain is easy, because then, as you said, $A$ would be an order of some number field in characteristic $0$, or $A$ would be finite in non zero characteristic. 
But what worries me is your hint (which I can't quite grasp) at the finite number of minimal primes of $A$ to reduce to that particular case, whereas Samuel feels obliged to go on with a technical inductive proof on the nilradical $N$ of $A$. More precisely, the induction bears on the exponent $s$ such that $N^s = (0)$. The starting step is $s=0$, i.e. $A$ is a reduced ring in which ($0$) is the intersection of finitely many prime ideals $P_i$'s, and so $A^*$ injects into the direct product of the $(A/P_i)^*$'s, which are of finite type according to the particular case. Next assume $s>1$ and consider the natural map $\phi : A \to A/N^{s-1}$. By the induction hypothesis $\operatorname{Im}\phi$ is finitely generated, and Samuel shows that $\ker\phi = 1+N^{s-1}$ and that the latter group is finitely generated. 
Finally, your original question, with the additional assumption that $A$ is reduced, has an affirmative answer, see P. Samuel, "A propos du théorème des unités", Bull. Sc. Math., 90 (1966), 89-96).
