THIS ANSWER IS WRONG (in that it claims the result in excess generality)
Suppose that $X$ is a $K(\pi,1)$ and a CW complex. Let $E$ be a universal covering space of $X$, so that $E$ is contractible and there is a free $\pi$-action on $E$ with orbit space $E/\pi\cong X$. (I really mean free and properly discontinuous -- what Hatcher's book calls a "covering space action".)
This gives $L(E)$ a $\pi$-action, again free, with orbit space $L(E)/\pi\cong L_0X$. And $L(E)$ has also $S^1$ acting, commuting with the $\pi$-action, in such a way that
On the other hand, a deformation retraction of $E$ to a point $p$ yields an $S^1$-equivariant deformation retraction of $L(E)$ to $L(p)$ (i.e. one that is compatible with the $S^1$-action), and this in turn yields a deformation retraction of $L(E)/S^1$ to $L(p)/S^1$.
So $L(E)/S^1$ is contractible and has a free action of $\pi$ with orbit space $L_0(X)/S^1$. This implies that $L_0(X)/S^1$ is a $K(\pi,1)$.
THE ERROR in the answer as written is in the assertion that the action of $\pi$ on $L(E)/S^1$ is free. But if we add the hypothesis that the group $\pi$ is torsionfree (which is true in the kinds of cases that the question is asking about) then this is all right: if $g\in\pi$ fixes the $S^1$-orbit of the loop $\gamma:S^1\to X$, then $g\circ \pi=\pi\circ R$ for some rotation $R:S^1\to S^1$; since the action of $\pi$ on $E$ is free (in the strong sense), this cannot happen unless $g$ and $R$ have (the same) finite order.