Yes, if you either assume also that $F$ is an isofibration or are content with proving that it's a Street fibration. And you only need pullbacks, not a terminal object.

Given $x\in C$ with $F(x)=b$ and a morphism $g:a\to b$ in $D$, let $g^*x$ be the pullback of the unit $\eta_x : x \to U(b) = UF(x)$ along $U(g) : U(a) \to U(b)$. Since $F(\eta_x)$ is an isomorphism (as $U$ is fully faithful) and $F$ preserves pullbacks, $F$ also maps the map $g^*x \to a$ to an isomorphism. And if $F$ is an isofibration, we can modify $g^*x$ to make this map to an equality. Finally, it's straightforward to prove that the other map $g^*x \to x$ is cartesian, using the universal property of the pullback.

Yes, if you either assume also that $F$ is an isofibration or are content with proving that it's a Street fibration. And you only need pullbacks, not a terminal object.

Given $x\in C$ with $F(x)=b$ and a morphism $g:a\to b$ in $D$, let $g^*x$ be the pullback of the unit $\eta_x : x \to U(b) = UF(x)$ along $U(g) : U(a) \to U(b)$. Since $F(\eta_x)$ is an isomorphism (as $U$ is fully faithful) and $F$ preserves pullbacks, $F$ also maps the map $g^*x \to U(a)$ to an isomorphism. And if $F$ is an isofibration, we can modify $g^*x$ to make this map to an equality. Finally, it's straightforward to prove that the other map $g^*x \to x$ is cartesian, using the universal property of the pullback.