Yes, but it's more complicated. First let me describe the special case where $\pi_1$ is trivial. If $G$ is such a 3-group, then $X = BG$ is a pointed connected simply connected space. It fits into a fibration
$$B^3 \pi_3 \to X \to B^2 \pi_2$$
which, as it turns out, can be delooped once into a fiber sequence
$$B^3 \pi_3 \to X \to B^2 \pi_2 \to B^4 \pi_3$$
reflecting the fact that the fibration above is classified by a map $f : B^2 \pi_2 \to B^4 \pi_3$, or equivalently by a cohomology class in $H^4(B^2 \pi_2, \pi_3)$. It's a classic result that this cohomology group is naturally isomorphic to the group of $\pi_3$-valued quadratic forms on $\pi_2$: this quadratic form corresponds to a homotopy operation $\pi_2 \to \pi_3$, represented by the Hopf fibration, which is a quadratic refinement of the Whitehead bracket $\pi_2 \times \pi_2 \to \pi_3$.
(As written, this is a classification of spaces with nontrivial $\pi_2$ and $\pi_3$. Delooping once, this is a classification of simply connected 3-groups. Delooping twice, this is a classification of grouplike braided monoidal groupoids, or said another way, braided 2-groups.)
The general case is messier. Now $X = BG$ fits into a fibration
$$\widetilde{X} \to X \to B \pi_1$$
where $\widetilde{X}$ is the universal cover of $X$, which is as above. Such fibrations are classified by actions of $\pi_1$ on $\widetilde{X}$ in a homotopy-theoretic sense, so now one has to compute the automorphisms of all possible $\widetilde{X}$, then maps from $\pi_1$ into these.
In the special case where $\pi_2$ is trivial, $X = BG$ now fits into a fibration
$$B^3 \pi_3 \to X \to B \pi_1$$
which implies that $X$ is classified by a pair consisting of an action of $\pi_1$ on $\pi_3$ and a cohomology class in $H^4(B \pi_1, \pi_3)$.