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NO, since the three-torus $T^3$ does not have this form.
EDIT if the OP really means a free product of $\mathbb{Z}$s, so the free group $F_k,$ then the answer is YES. It is a fact (see Hempel's book, chapter 7) that every splitting of the fundamental group of $M^3$ as a free product comes from a connected sum decomposition. On the other hand, a prime three manifold is either a $K(\pi, 1)$ or $S^2 \times S^1.$ In the former case, its fundamental group cannot be $\mathbb{Z}$
No, since the three-torus $T^3$ does not have this form.