The answer is no. Here is a counterexample. Let $p$ be a prime.

Let $F = \mathbb Z / p \otimes_{\mathbb Z}^L (-) : D(\mathbb Z) \to D(\mathbb Z)$, and let $\phi = p : \mathbb Z \to \mathbb Z$. Take $Z,Z'$ to be appropriate shifts of $\mathbb Z$ and choose $\psi$ so that $Fib(\psi) = \mathbb Z / p^2$ (after all, there is a short exact sequence $0 \to \mathbb Z / p \to \mathbb Z / p^2 \to \mathbb Z / p \to 0$). Since $p = 0$ on $\mathbb Z / p$ -modules, but $p \neq 0$ on $\mathbb Z / p^2$ or its dual, you get a counterexample.

The answer is no. Here is a counterexample. Let $p$ be a prime.

Let $F = \mathbb Z / p \otimes_{\mathbb Z}^L (-) : D(\mathbb Z) \to D(\mathbb Z)$, and let $\phi = p : \mathbb Z \to \mathbb Z$. Take $Z' = \mathbb Z, Z = \Sigma \mathbb Z$ to be appropriate shifts of $\mathbb Z$ and choose $\psi : \mathbb Z / p \to \Sigma \mathbb Z / p$ so that $Fib(\psi) = \mathbb Z / p^2$ (after all, there is a short exact sequence $0 \to \mathbb Z / p \to \mathbb Z / p^2 \to \mathbb Z / p \to 0$). Since $p = 0$ on $\mathbb Z / p$ -modules, but $p \neq 0$ on $\mathbb Z / p^2$ or its dual, you get a counterexample.