The Evans conjecture ( which was proved later by [Smetaniuk]()) states that for any $n$, if at most $n-1$ entries of a partial $n\times n$ latin square are filled,  it can be completed to the full latin square. 

My question pertains to whether this is applicable to symmetric (or commutative) latin square? That is, given $n\times n$ latin square, with at most $n-1$ prefilled entries ( which are symmetric with respect to diagonal), can it be completed to a symmetric latin square? Note that this corresponds to pre- total coloring of a complete graph of order $n$ with at most $n-1$ previous filled colors.  Any hints? Thanks beforehand.