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.