I will show that this is not the case for $\mathbb{F}_9$$F=\mathbb{F}_9$. The proof generalize to any $\mathbb{F}_{p^k}$, with $k >1$.
I'm starting from the observation that: $\mathbb{F}_9 \simeq \mathbb{Z}[i]/(3) \simeq \mathbb{Z}[\sqrt{2}]/(3)$. where I'm using the isomorphism that identifies $i$ and $\sqrt{2}$ to identify $\mathbb{Z}[i]/(3)$ and $\mathbb{Z}[\sqrt{2}]/(3)$.
I'm now considering $A= \mathbb{Z}[i,\sqrt{2}]$. It has a surjective map $\phi:A \to \mathbb{F}_9$ induced by the two maps $\mathbb{Z}[i] \to \mathbb{F}_9$ and $\mathbb{Z}[\sqrt{2}] \to \mathbb{F}_9$ above.
I can localize $A$ at $\ker \phi$ to make it an object of your category$C_F$.
So, in the category under consideration$C_F$ I have a diagram:
$$ \mathbb{Z}[i]_{(3)} \to A_{\ker \phi} \leftarrow \mathbb{Z}[\sqrt 2]_{(3)} $$
If there was an initial object $B$ in your category$C_F$, itits unique map to $A_{\ker \phi}$ should factor thoughfactors through both $\mathbb{Z}[i]_{(3)} $ and $\mathbb{Z}[\sqrt 2]_{(3)} $$\mathbb{Z}[\sqrt 2]_{(3)}$ hence, it should factor through their intersection, but thein $A_{\ker \phi}$. But this intersection is reduced to $\mathbb{Z}_{(3)}$ whose, so we should have a map $B \to \mathbb{Z}_{(3)}$ compatible with the map back to $\mathbb{F}_9$, but as the map $\mathbb{Z}_{(3)} \to \mathbb{F}_9$ is not surjective, so no object of your category can factor through itand the map $B \to \mathbb{F}_9$ needs to be, we have a contradiction.