There is no such thing as "injective" or "surjective" morphisms; these conditions only make sense in the presence of a forgetful functor to $\text{Set}$.
What is true is that exact functors preserve monomorphisms and epimorphisms, and this is because
- in an abelian category, monomorphisms are precisely the morphisms with zero kernel and epimorphisms are precisely the morphisms with zero cokernel,
- exact functors preserve kernels and cokernels, and
- exact functors preserve zero.
Edit: The answer to the revised question is no. The next few paragraphs explain an abstract argument for finding large classes of counterexamples, and towards the end give a specific counterexample, which is easier to understand.
For starters, let $F$ be the identity functor between two copies of the same abelian category $A$ which are concretized with respect to two different forgetful functors $U_1, U_2 : A \to \text{Set}$. If $f$ is a morphism in $A$, call it $U_1$-surjective if $U_1(f)$ is surjective and $U_2$-surjective if $U_2(f)$ is surjective. We are looking for examples of $A, U_1, U_2$ such that $U_1$-surjectivity does not imply $U_2$-surjectivity, which will turn out to be a bit easier to find than examples involving injectivity.
However, if $F : C \to D$ is a counterexample to the question about surjectivity, then $F^{op} : C^{op} \to D^{op}$ is a counterexample to the question about injectivity, provided that the forgetful functors $U_C, U_D : C, D \to \text{Set}$ are replaced by the composites of their opposites
$$U_C^{op}, U_D^{op} : C^{op}, D^{op} \to \text{Set}^{op}$$
with the functor $\text{Hom}(-, 2) : \text{Set}^{op} \to \text{Set}$.
We can construct a general class of counterexamples as follows. Suppose $U_1, U_2$ take the form $\text{Hom}(u_1, -)$ and $\text{Hom}(u_2, -)$ for two generators $u_1, u_2$ of $A$, and that $u_1$ is projective but $u_2$ is not. This means that $U_1$ is exact and faithful, and hence that a morphism is $U_1$-surjective iff it's an epimorphism (exercise).
It also means that $u_2$ admits a nontrivial extension
$$0 \to a \to b \xrightarrow{f} u_2 \to 0$$
for some objects $a, b \in A$. The morphism $f$ is an epimorphism, hence $U_1$-surjective, but it cannot be $U_2$-surjective: applying $\text{Hom}(u_2, -)$ to this short exact sequence produces the Ext long exact sequence
$$0 \to \text{Hom}(u_2, a) \to \text{Hom}(u_2, b) \xrightarrow{U_2(f)} \text{Hom}(u_2, u_2) \xrightarrow{\partial} \text{Ext}^1(u_2, a) \to \cdots$$
where $U_2(f)$ cannot be surjective because the connecting homomorphism $\partial$ is not zero: it necessarily maps $\text{id} \in \text{Hom}(u_2, u_2)$ to the element of $\text{Ext}^1(u_2, a)$ classifying the above extension. From here it suffices to find an abelian category $A$ and two generators $u_1, u_2$ of it, one of which is projective and one of which is not.
This state of affairs is easy to arrange; arguably the simplest example is $A = \text{Ab}, u_1 = \mathbb{Z}, u_2 = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, so that
$$U_1(-) = \text{Hom}(\mathbb{Z}, -) : \text{Ab} \to \text{Set}$$
is the usual forgetful functor, but
$$U_2(-) = \text{Hom}(\mathbb{Z}, -) \oplus \text{Hom}(\mathbb{Z}/2\mathbb{Z}, -) : \text{Ab} \to \text{Set}$$
sends an abelian group to its underlying set times its subgroup of elements of order dividing $2$. So a morphism $f : x \to y$ of abelian groups is $U_1$-surjective iff it's surjective, but $U_2$-surjective iff it is both surjective and surjective on elements of order dividing $2$. This means that e.g. the quotient map $\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ is surjective but not $U_2$-surjective.
