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Anton Petrunin
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As it was shown by Igor, there is no univeral bound on number of such geodesics. Let me show that the number can not be infinite.

Assume it is possible to get infinite number of such geodesics, say $\gamma_n$, $n\in\mathbb N$. Note that the geodesics $\gamma_i$ for $i\le n$ cut $\mathbb S^2$ into surfaces with geodesic boundaries. By Gauss--Bonnet formula most of these surfaces are cylinders.

By passing to a subsequence, we can assume that for each $n$, the geodesics $\gamma_i$ for $i\le n$ cut $\mathbb S^2$ into two discs and cylinders between $\gamma_i$ and $\gamma_{i+1}$.

Denote by $\gamma_\infty$ the limit of $\gamma_n$ as $n\to\infty$. Note that this limit is defined and the limit geodesic $\gamma_\infty$ is stable.

Given a point $p$ near $\gamma_\infty$ denote by $\ell(p)$ the length of mimimal geodesic loop based at $p$ which goes sufficiently close to $\gamma_\infty$. Note that $\ell$ is an analytic function and its derivatives vanish on $\gamma_\infty$. It follows that $\ell\equiv 0$; i.e. $\gamma_\infty$ lies in a one parameter family of closed geodesics which sweep a neighborhood of $\gamma_\infty$.

Pass to the analytical extension of this one parameter family, lets denote it by $\xi_\tau$. Note that the geodesics in the family stay simple and disjoint locally. Globally, it only may happen that $\xi_0=\xi_c$ for some parameter $c\ne0$. Moreover, since the surface is compact it actually happens. In this case the surface is a total space of a circle bundle, a contradiction.

Anton Petrunin
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