Are all Orbits in Schwarzschild geometry planar? My question is if all Orbits in Schwarzschild geometry are planes?
A common way to argue that there are planar orbits is by noticing that $\eta= \frac{\pi}{2}$ is a solution of the corresponding equation of motion and by sphereical symmetry we can without loss of generality consider those Orbits.
Now I asking my self if there is a proof which states that all possible solutions to the equation of motion have planar Orbits?
 A: It's been a long time since this question has been asked, but perhaps someone might still strand here looking for an answer, so let me give it a try. 
I think the crucial point here is the uniqueness of geodesics. Given initial conditions, i.e. a point in your manifold and and initial velocity vector, there always exists a unique geodesics satisfying these initial conditions. Now you can easily argue that, given a point and a velocity vector, there exists a planar geodesic satisfying these initial conditions (by performing suitable rotations, as already mentioned in another answer). But then by the uniqueness of geodesics, this already exhausts all possible geodesics in the geometry. There cannot be more. Hence all geodesics are planar.
An alternative way to see that all orbits must be planar is to realize that first of all $p_\phi$ is conserved along geodesics, where $p_\mu\equiv g_{\mu\nu}\dot x^\nu$, (so say $p_\phi = k$ for some constant $k$) and that the quantity
$$L^2 = p_\theta^2 + \dfrac{p_\phi^2}{\sin^2\theta } = r^4\dot\theta^2 + \dfrac{k^2}{\sin^2\theta }$$
is conserved along geodesics as well. Then the crucial thing to note is that this function $L^2$ attains a strict minimum $L^2 = k^2$ at $\theta=\pi/2,\, \dot\theta= 0$ in the sense that $\theta=\pi/2,\, \dot\theta= 0$ is the only situation in which $L^2$ has this value. For all other values, both terms occurring in $L^2$ become strictly larger (except for $\theta =-\pi/2$, $\dot\theta =0$ but in order to reach these values $\theta$ first have to go through other values for which $L^2$ does grow.) Since $L^2$ is conserved it is not possible to go over into other values and hence if $\theta=\pi/2,\, \dot\theta= 0$ is satisfied at some point, then $\theta=\pi/2,\, \dot\theta= 0$ will be satisfied along all of the geodesic. 
This argument proves that all geodesics starting along the 'standard' equator are planar. But, again, by spherical symmetry, every single geodesic can be viewed as a geodesic starting along a 'standard' equator if we rotate the coordinate system suitably. Hence all geodesics must be planar.
Note: to clear any confusion about notation, my coordinates in the discussion above are such that the Schwarzschild metric reads
$$ \text{d} s^2 =  -c^2\left(1-\dfrac{2 G M}{c^2 r}\right)\text{d} t^2 + \left(1-\dfrac{2 G M}{c^2 r}\right)^{-1} \text{d}r^2 + r^2\left(\text{d} \theta^2 +\sin^2\theta 
 \text{d}\phi^2\right). $$
A: by a rotation, any arbitrary initial conditions can be transformed to planar ones, in exactly the way that you said, and then you can undo the transformation after solving the EOM.  I don't see what more of a proof you need. 
