Since you are working with a spin-1/2 particle, it is more conventional to use $\sigma_{xyz}$ than $S_{xyz}$. This makes it clear that your Hamiltonian will be a matrix proportional to $\sigma_x$:
$$
H=-\gamma B\sigma_x=-\gamma B\left(\begin{array}{cc} 0&1 \\ 1&0\end{array}\right)\, \qquad B_x=B \tag{1}
$$
in a basis where $\sigma_z$ is diagonal and where $\vert +z\rangle = (1,0)^T$.
The technical difficulty with your approach is that you get a pair of coupled differential equations. Uncoupling these involves diagonalizing a matrix, i.e. finding the eigenstates of $H$. You want these eigenstates because it is these eigenstates of $H$ that have simple time evolution,
If you stay away from the differential equations and focus on the matrix approach as per (1), the solution is achieved in three steps. First, find the eigenstates of $H$. Second, express your initial state as a combination of eigenstates of $H$, and third time-evolve your initial state by time-evolving each eigenstates of $H$ in the linear combination giving your initial state.
You can proceed as your suggest, i.e. “rotate” everything so that you now quantize about $\hat x$. In this case, you would need to find a transformation $T$ so that $T\sigma_x T^{-1}$ is diagonal and express your initial state $\vert +z\rangle$ as $T\vert +z\rangle$. You do not gain much by doing this as you are required to find $T$ - which you will need to find anyways to get the eigenstates of $\sigma_x$ in the $z$-basis.
There is a final trick, related to directly solving your differential equations. Your differential equation approach should get you a pair of equations of the type
$$
\dot{\alpha}(t)= c_1 \beta(t)\, ,\qquad \dot{\beta}(t)= c_2\alpha(t)\, , \tag{2}
$$
with $c_1$ and $c_2$ numbers related to $H$. The form of Eq.(2) is specific to your Hamiltonian. From (2) take the time-derivative of the first and use the second to a second order equation for $\ddot{\alpha}$, which you should be able to solve given your initial conditions.
