Yes, this is true.
First of all, the map $\newcommand{\MM}{\mathcal{\overline M}_{g,n}}A^1(\MM) \to H^2(\MM)$ is an isomorphism (say with $\mathbf Q$-coefficients on both sides), so by Poincaré duality it will be enough to prove that $H^{6g-6+2n-2}(\MM)$ is spanned by classes of F-curves.
We do this using the spectral sequence associated to a filtration. Filter $\MM$ by taking $X_k$ to be the union of all strata of dimension $\leq k$ in the stratification by topological type, so we have a chain of closed subvarieties $$ \emptyset = X_{-1} \subset X_0 \subset \cdots \subset X_{3g-3+n} = \MM.$$ I always find it easier to think about the Poincaré dual spectral sequence in compactly supported cohomology, which reads $$ E_1^{pq} = H^{p+q}_c(X_p \setminus X_{p-1} ) \implies H^{p+q}_c(\MM).$$ Each $X_p \setminus X_{p-1}$ is a disjoint union of strata, each of the form $\prod \mathcal M_{g',n'}$ (possibly divided by a finite group). We want to show that $H^2_c(\MM)$ is spanned by F-curves. Now $H^2_c(X_0)$ obviously vanishes. $H^2_c(X_1 \setminus X_0)$ is given by the fundamental classes of all 1-dimensional strata. These are by definition the F-curves. Finally I claim that also $H^2_c(X_k \setminus X_{k-1})$ vanishes for all $k > 1$. This is (for instance) an easy consequence of the virtual cohomological dimension of $\mathcal M_{g,n}$.