I wouldn't say we "prefer" the eigenvalues to the singular values in the symmetric/Hermitian case. They are effectively the same thing! Or, more accurately, the singular values are nothing more than the absolute values of the eigenvalues.
To see why, let $X$ be any Hermitian matrix. Its Schur (eigenvalue) decomposition is $X=Q\Lambda Q^H$ where $\Lambda$ is diagonal, and $Q^HQ=QQ^H=I$. Its singular value decomposition, on the other hand, is $X=U\Sigma V^H$, where $\Sigma$ is diagonal and nonnegative, and $U^HU=UU^H=I$ and $V^HV=VV^H=I$.
Alternatively, we can write these decompositions in dyadic form:
$$X=\sum_{i=1}^n \lambda_i q_i q_i^H = \sum_{i=1}^n \sigma_i u_i v_i^H$$
where $q_i$, $u_i$, and $v_i$ are the $i$th columns of $Q$, $U$, and $V$, respectively, and $\lambda_i$ and $\sigma_i$ are the $i$th diagonal elements of $\Lambda$ and $\Sigma$, respectively.
In this latter form, it is simple to see that, given the Schur decomposition,
$$\sigma_i = |\lambda_i|, \quad u_i = q_i, \quad v_i = \begin{cases} v_i & \lambda_i \geq 0 \\ -v_i & \lambda_i < 0 \end{cases}
\qquad i=1,2,\dots, n$$
gives us a valid singular value decomposition. Our choice to apply the sign flipping to $v_i$ is arbitrary.
Be assured that in practice, this is precisely what you should be doing to compute the singular value decomposition of a Hermitian matrix. The symmetric eigenvalue problem is well-conditioned, and there is no numerical advantage to applying the a nonsymmetric SVD to a Hermitian matrix. In fact, the computation of the singular values of a non-Hermitian matrix $A$ are sometimes computed by applying a symmetric eigenvalue algorithm to the Hermitian matrix
$$\begin{bmatrix} 0 & A \\ A^T & 0 \end{bmatrix}$$
I'll leave it as an exercise for you to prove that this works :-)
EDIT: DGrady's comment alerted me to the fact that I didn't read the second half of the question correctly. In it, he was comparing the merits of computing the SVD of $X$ versus the eigenvalue decomposition of $C=XX^H$. In this case, it is indeed going to be a bit better to compute the SVD of $X$ directly.
Best Answer
You can compute a complete orthogonal decomposition using two QR decompositions. See http://www.netlib.org/lapack/lug/node43.html