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 Matthew Henderson.
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.