Yes, this is true. There are various ways to prove this. Here's the shortest argument I can think of. One direction is easy to prove, so let's prove the other direction.

Let $U \colon \mathbf{sSet}^+ \to \mathbf{sSet}$ denote the functor that forgets markings. We will use that the restriction of this functor to the full subcategory of fibrant objects in $\mathbf{sSet}^+$ (i.e. the naturally marked quasi-categories) is fully faithful and preserves cofibrations, fibrations, and weak equivalences. All of these properties are easy to prove.

Now, let $f \colon A \to B$ be a morphism of naturally marked quasi-categories, and suppose that $U(f) \colon U(A) \to U(B)$ is an isofibration of quasi-categories. We want to prove that $f$ is a fibration in the model structure on $\mathbf{sSet}^+$. Let $f = p\circ j$ be a factorisation of $f$ into a trivial cofibration $j \colon A \to C$ followed by a fibration $p \colon C \to B$ in $\mathbf{sSet}^+$. (Note that $C$ is fibrant, since $B$ is fibrant and $p$ is a fibration.) By the above preservation properties of $U$, $U(f) = U(p) \circ U(j)$ is a factorisation of $U(f)$ into a trivial cofibration followed by a fibration in the Joyal model structure on $\mathbf{sSet}$. But $U(p)$ is a fibration, so it is has the RLP wrt $U(j)$, and hence is a retract of $U(p)$ in $\mathbf{sSet}$. By the above fully faithfulness property of $U$ (restricted to the fibrant objects of $\mathbf{sSet}^+$), it follows that $f$ is a retract of $p$ in $\mathbf{sSet}^+$. Hence $f$ is a fibration in $\mathbf{sSet}^+$, since $p$ is.