To amplify, let me point out that it is not even true that an additive functor that preserves epimorphisms/surjections also preserves cokernels. For example, let $\mathcal{C}$ be the full subcategory of $\textbf{Ab}$ spanned by the torsion-free abelian groups. This category, perhaps unexpectedly, is an additive category with kernels and cokernels – but is not an abelian category. Indeed, in $\mathcal{C}$, the cokernel of $2 \cdot {-} : \mathbb{Z} \to \mathbb{Z}$ is $0$, so the inclusion $\mathcal{C} \hookrightarrow \textbf{Ab}$ is an additive functor that preserves surjections but not cokernels (or even epimorphisms in general!). In fact, $\mathcal{C} \hookrightarrow \textbf{Ab}$ is even a right adjoint, and so is left exact in particular. (When I say left/right exact, I always mean a functor that preserves finite limits/colimits.)
However, what is true is that a left exact functor $F : \mathcal{A} \to \mathcal{B}$ that preserves (normal) epimorphisms will be exact, provided $\mathcal{A}$ and $\mathcal{B}$ are both abelian. Indeed, since $\mathcal{A}$ and $\mathcal{B}$ are additive, to show that $F$ is right exact it is enough to show that it is additive and preserves cokernels. Now, it is known that a functor $\mathcal{A} \to \mathcal{B}$ that preserves finite products is automatically additive; but a left exact functor preserves finite products, so is an additive functor in particular. Consider a sequence of morphisms
$$0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0$$
in $\mathcal{A}$, and suppose $A' \to A$ is kernel of $A \to A''$ and $A \to A''$ is the cokernel of $A' \to A$. Since $F$ preserves kernels, we get an exact sequence
$$0 \longrightarrow F A' \longrightarrow F A \longrightarrow F A''$$
in $\mathcal{B}$, and since $A \to A''$ is a (normal) epimorphism, we can extend the above to a short exact sequence in $\mathcal{B}$:
$$0 \longrightarrow F A' \longrightarrow F A \longrightarrow F A'' \longrightarrow 0$$
Thus, $F$ preserves cokernels of normal monomorphisms. In general, if we have a morphism $X \to A$ in $\mathcal{A}$, we can factor it as $X \to A' \to A$ where $X \to A'$ is the cokernel of the kernel of $X \to A$, and it is not hard to show that the cokernel of $A' \to A$ is also the cokernel of $X \to A$. Since $F$ preserves all kernels and also cokernels of (normal) monomorphisms, $F$ preserves this factorisation, and therefore the cokernel of $F A' \to F A$ is also the cokernel of $F X \to F A$. However, because $\mathcal{A}$ is an abelian category, $A' \to A$ itself is a (normal) monomorphism, so $F$ preserves its cokernel. Thus $F$ actually preserves all cokernels and is therefore right exact.
Dually, of course, a right exact functor between abelian categories is exact if and only if it preserves (normal) monomorphisms. This explains the classical fact that $M$ is flat if and only if $- \otimes_R M$ preserves injective homomorphisms: $- \otimes_R M$ is a left adjoint and so right exact in particular.
Here is an minimal example satisfying both conditions at the same time, maybe not super interesting in itself but there we go.
Let $C$ and $D$ be two subcategories of the category of sets (so in particular they are concrete categories), defined by:
the objects of $C$ are $\{1\}$ and $\{2\}$, and the only morphisms are the identities, and the only map $\{1\}\to \{2\}$ (we do not take the map $\{2\}\to \{1\}$;
the objects of $D$ are $\{1,2\}$ and $\{3,4\}$, and the only morphisms are the identities, and the constant map $\{1,2\}\to \{3,4\}$ with value $4$.
Then those two categories are equivalent (they are both equivalent to the abstract category $\bullet\to \bullet$), but the non-identity map in $C$, which is a bijection, is sent to the map in $D$ which is neither injective nor surjective.
Best Answer
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
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.