Consider in $\Bbb S^2$ the symplectic form $\omega \in \Omega^2(\Bbb S^2)$ given by $\omega_p(v,w) = \langle p, v \times w\rangle$. Fixed $z \in \Bbb R^3$, we define $h_z\colon \Bbb S^2 \to \Bbb R$ by $h_z(p) \doteq \langle p,z\rangle$. I computed the Hamiltonian vector field of $h_z$ as $X_{h_z}(p) = z \times p$.
I would like to find the integral curves of $X_{h_z}$. This means solving the system $\alpha'(t) = z \times \alpha(t)$. If $z = (a,b,c)$ and $$Z = \begin{pmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{pmatrix},$$we want to solve $\alpha'(t) = Z\alpha(t)$. Let's put the initial condition $\alpha(0) = p_0$ to ensure uniqueness of solution. Then $$\alpha(t) = \exp(tZ)p_0,$$and we use Rodrigues' rotation formula to get $$\alpha(t) = \left({\rm Id} + \frac{\sin \|tz\|}{\|tz\|}Z + \frac{1-\cos\|tz\|}{\|tz\|^2}Z^2\right)p_0.$$I'm having trouble visualizing this solution even in simple cases. We can assume that $p_0 = (0,0,1)$ since rotations are symplectomorphisms. If for example $z = (0,0,1)$, then $\alpha$ degenerates to a point. If $z = (0,1,0)$, then $\alpha(t)$ has the form $(\ast,0,\ast)$ and so is a pre-geodesic. What happens in the general case?
My sign convention is that if $(M,\omega)$ is a symplectic manifold and $f \in \mathcal{C}^\infty(M)$, then $X_f$ satisfies $\iota_{X_f}\omega = {\rm d}f$.
Let me present four easier points of view which are intertwined with visualization, but easily put into calculations.
Your computation of the Hamiltonian vector field could be used directly to visualize it as in the second drawing which comes below ($z \times p$ is a vector of constant norm along the parallel passing through $p$ with respect to $z$, and it is tangent to said parallel. This gives an explicit solution: the parallel parametrized as such to have constant speed $\Vert z \times p\Vert$, and in the correct direction given by the sign conventions), but it may be a little ellusive to understand what is going on directly.
We can assume $z=(0,0,1)$ and $h$ is the height function after applying a rotation. This is not so important, but it is easier to present a drawing (which I will later on).
Since the north pole and the south pole clearly are stationary points since the height function has a maximum (resp. minimum) there, it suffices to analyse the rest. Therefore, we can use the fact that the horizontal projection of the cylinder on the sphere is a symplectomorphism (exercise!), and the pull-back of the height function under this map is obviously the height function on the cylinder. On the cylinder, it is clear (exercise!) that the integral curves are the horizontal circles. So the integral curves on the sphere will also be the horizontal circles (with different speeds corresponding to the fact that the speeds of the horizontal circles on the cylinder will transfer to slower speeds when sent via the symplectomorphism). The direction of those circles depends on the orientations you take.
The sphere $S^2$ has an obvious almost complex structure given by $J_px=p \times x$ (or $x \times p$, again orientation issues and sign conventions etc). This is just a rotation of $\pi/4$. It is easy to verify (exercise!) that $J$ is a compatible almost complex structure with the metric on the sphere, i.e., $$\omega(\cdot,J\cdot) =\langle \cdot, \cdot \rangle$$ and $$\omega(J\cdot,J\cdot)=\omega(\cdot, \cdot).$$ Thus, the Hamiltonian vector field is given by $-J\nabla h$. And it is clear what that is, and what the orbits are:
Since $\nabla h$ is the vector field
it follows that $X_h$ is
PS: The drawing is a little off, as the vectors should be larger in the equator.
For a given integral path $\gamma$ of the Hamiltonian flow, $H \circ \gamma$ is constant. Thus, $\gamma$ must lie in a parallel of the sphere. Using a fragment of the other points of view, you can see that the speed must be constant on each integral path, thus the solutions are clear (computing the actual speed in terms of the height is easy).