There's actually an extremely nice way to uncover the Bloch sphere representation for any density operator. (Pure states are just a special case.)
Definition.
I'm not sure how your (lecturer? book? other learning source?) defines the Bloch sphere, but the definition that makes the most sense from a fundamental perspective is that, for any density operator ρ, the point on (or inside) the Bloch sphere corresponding to ρ is the vector (rx, ry, rz) such that
$$ \rho = \tfrac{1}{2}( I + r_x X + r_y Y + r_z Z )$$
where $I$ is the identity and $X, Y, Z$ are the (other) 2×2 Pauli operators.
Proof sketch.
It's easy to show that the operators $I, X, Y, Z$ are linearly independent (what linear combinations of them add to the zero operator?) and are Hermitian (each is equal to it's own conjugate-transpose). From this you can show that they span the set of all 2×2 Hermitian operators; and as they are linearly independent, they're actually a basis set for those operators. So any density operator — which is also Hermitian — will decompose into $I, X, Y, Z$ in a unique way. (It's possible to show that it's coefficient in $I$ is always ½ by considering the trace. Do you see how?)
Answer.
You should try to prove the things I've said above — it isn't hard, and it's using math that will be useful to you later anyway — but for the problem of finding the Bloch sphere representation, all you need to do is solve for (rx, ry, rz) in the equation above.
If you like, you can even obtain these coefficients by a simple formula. (Hint: what is the trace of the product of two different matrices chosen from $I,X,Y,Z$? What does this mean for $\mathrm{tr}(\rho P)$ for $P \in \{I,X,Y,Z\}$?)
Another remark —
In the future, you don't have to really do any work to find the eigenvalues of a diagonal matrix $D$. It's easy to show that he standard basis vectors $\mathbf e_j = [\; 0 \; \cdots \; 0 \;\; 1 \;\; 0 \; \cdots \; 0 \;]^\top$ are eigenvectors for any diagonal matrix, and that the eigenvalues are exactly the coefficients on the diagonal (with multiplicity given by how often each is repeated). It's also easy to show that $D - \lambda I$ is invertible for any other $\lambda$, so that these are all the eigenvalues.
The rotation operator normalization you have chosen, $ R_{\boldsymbol n}(\theta)=e^{-i\theta \boldsymbol \sigma\cdot \boldsymbol n/2} $, means that for a rotation by $2\pi$ about the $z$ axis,
$$ R_{\hat z}(2\pi)=e^{-i\pi \sigma_z} =(-1)^{\sigma_z}=-1,$$
because $(-1)^1=(-1)^{-1}=-1$. Thus, your rotation operator has indeed rotated you back to the same point on the Bloch sphere (because the operator is scalar) but you have accumulated a global phase. There isn't, strictly speaking, an error, because global phases can be ignored.
On the other hand, this 'global' phase is tremendously useful if you are in contact with some other degree of freedom. It is an example of a geometric phase (i.e. if you rotate around some path in the Bloch sphere and return to the same path, you will accumulate a phase of
$$\text{(area enclosed by the path in steradians)}/2,$$
where the total area to be had is of course $4\pi$). It is the foundation of what are called phase gates, which are very popular basic entangling gates between two qubits. The idea there is that you perform a controlled rotation: leave qubit A alone if B is in the down state, and rotate it once around the Bloch sphere if B is in the up state. Either way, you come back to the same state, but in one of the two paths you have accumulated a phase, which can be used to generate entanglement.
Best Answer
Here I give some additional analysis to Craig Gidneys answer. The Hadamard gate indeed can be subdivided into two individual rotations, as suggested in the initial question. However, every single qubit gate can also be described by a single rotation on the Bloch sphere. For the Hadamard gate, this is a rotation around the axis (rho, elev, azim) = $(1, \frac{\pi}{4}, 0)$. The rotation axis is drawn as a black arrow in the image.
Shown is the evolution of the initial state $|0\rangle$ during two sequentially applied Hadamard gates. The red state vectors show the evolution during the first Hadamard gate, transferring state $|0\rangle$ to state $|x\rangle$ (in the image only x). The blue state vectors then show the evolution during the second Hadamard gate, transferring state $|x\rangle$ back to state $|0\rangle$. This geometric representation shows for one exemplary initial state, why $H^2 = I$.
The image can be resimulated with python and the toolbox qutip.