It's well known that it isn't true. Here is a counterexample: (1,1;0 1)$\begin{pmatrix}1&1\\0&1\end{pmatrix}$. This is not hermitian, but it has two real eigenvalues +1,+1. This example is not diagonalizable, so it isn't so interesting.
A diagonalizable example is easy to construct too, if the eigenvectors are not orthogonal to each other. Consider the matrix (100,3;-2,234)$\begin{pmatrix}100&3\\-2&234\end{pmatrix}$. This matrix has two real eigenvalues close to 100 and 234, since the small perturbation of the eigenvalue equation doesn't change the discriminant. But the matrix is not symmetric, so it is not Hermitian. In this case, you can define a different metric on the vector space, a different definition of orthogonal, that makes the matrix Hermitian. This is easy-- the matrix is diagonal in it's Eigenbasis, with real eigenvalues, if you declare that this basis is orthonormal, then the matrix becomes Hermitian.
If you have a diagonalizable matrix with real eigenvalues $E_i$, and the eigenvectors $V_i$ are orthogonal and form a complete set,
Then the matrix is given by
$$ E_i \bar{V}_i^j V_i^k $$
This reconstructs a Hermitian matrix from the list of orthogonal real eigenvalues. A proper statement is that a diagonalizable matrix with real eigenvalues and a basis of eigenvectors defines a metric on the complex vector space where it becomes Hermitian. The proof is to declare that all the eigenvectors have zero inner product, and some positive norm.
In the subject of PT symmetric quantum mechanics, this construction defines the metric on Hilbert space from the energy eigenstates.
