No, when it comes to symmetric latin squares it is no longer true that as many as $n-1$ cells can be prescribed unconditionally. This is explained in the Ph.D. thesis of <A HREF="https://mjh-phd.netlify.app/sec110.html">Matthew Henderson.</A>

> The key point here is that in a symmetric latin square, precisely
> because of the symmetry, every symbol $\sigma$ occurs an even number
> of times in cells outside of the main diagonal. Therefore, as every
> symbol $\sigma$ occurs $n$  times in total it follows that the number
> of cells of the main diagonal containing symbol $\sigma$ is congruent
> to $n$ modulo 2. A partial latin square can be incompletable because
> there are more symbols which occur on the main diagonal a number of
> times (zero included) incongruent to $n$  modulo 2 than there are
> empty cells on the main diagonal.    
> If this obstruction does not occur the diagonal is called "admissible". For $n$ odd a diagonal is admissible if and only if no symbol occurs more than once on it.

The generalization of Evans theorem to symmetric latin squares by <A HREF="https://www.academia.edu/20306267/Symmetric_latin_square_and_complete_graph_analogues_of_the_evans_conjecture">Andersen and Hilton</A> is that the symmetric latin square is completable if the diagonal is admissible and at most $n-1$ cells in total are prescribed.

In response to the comment by Richard Stanley: it is not sufficient to only count cells on or above the main diagonal, as this counter example shows. The diagonal is admissible ($n=5$ is odd and no symbol appears more than once on the diagonal), and the number of prescribed cells on or above the main diagonal is $n-1=4$, and yet the latin square is not completable.

<IMG SRC="https://ilorentz.org/beenakker/MO/latin_square_symmetric.png"/>