Yes.
First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.
If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$. Then $\chi(\partial M)=k$. But $\chi(\partial M)=2\chi(M)$ for any compact 3-manifold so $k$ is even.
By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $S$ be a 2-sphere embedded in $N$ which separates two components of $\partial N$ from the other $k-2$ components. Let $V$ and $W$ be the closures of the components of $N\setminus S$ such that $\partial V$ consists of $k-1$ 2-spheres and $\partial W$ consists of three 2-spheres.
Suppose $p(S)$ is a projective plane. Then $p(W)$ is a compact 3-manifold with $\chi(\partial p(W))=3$, a contradiction. Hence $p(S)$ is a 2-sphere, and $p(S)$ is separating in $M$.
Note that $M$ is prime. Otherwise $\pi_1(M)$ would split as a free product. But non-trivial free products are infinite (and non-abelian). So we again use the Poincaré Conjecture to infer that any simply-connected summand must be the 3-sphere.
Thus $p(S)$ must bound a 3-ball in $M$. Hence $p(V)$ is a 3-ball. Then $V$ is also a 3-ball, which implies that $k=2$. Therefore $N$ is homeomorphic to $S^2\times I$, and it follows that $M$ is homeomorphic to $F\times I$.
