For the case $n=8$, with the precoloring you describe the completion you give is indeed **unique**. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one. 


[![Colored C_8][1]][1]


For the case $n=10$, consider the pre-colored $K_{10}$
$$\left(\begin{array}{rrrrrrrrrr}
X &   &   &   &   & 1 & 8 & 4 & 9 & 5 \\
  & X &   &   &   &   & 2 & 9 & 5 & 1 \\
  &   & X &   &   &   &   & 3 & 1 & 6 \\
  &   &   & X &   &   &   &   & 4 & 2 \\
  &   &   &   & X &   &   &   &   & 7 \\
1 &   &   &   &   & X &   &   &   &   \\
8 & 2 &   &   &   &   & X &   &   &   \\
4 & 9 & 3 &   &   &   &   & X &   &   \\
9 & 5 & 1 & 4 &   &   &   &   & X &   \\
5 & 1 & 6 & 2 & 7 &   &   &   &   & X
\end{array}\right)$$ 
This can be completed in $77$ ways, for example 
$$\left(\begin{array}{rrrrrrrrrr}
X & 6 & 7 & 3 & 2 & 1 & 8 & 4 & 9 & 5 \\
6 & X & 8 & 7 & 3 & 4 & 2 & 9 & 5 & 1 \\
7 & 8 & X & 5 & 4 & 2 & 9 & 3 & 1 & 6 \\
3 & 7 & 5 & X & 9 & 8 & 6 & 1 & 4 & 2 \\
2 & 3 & 4 & 9 & X & 5 & 1 & 6 & 8 & 7 \\
1 & 4 & 2 & 8 & 5 & X & 3 & 7 & 6 & 9 \\
8 & 2 & 9 & 6 & 1 & 3 & X & 5 & 7 & 4 \\
4 & 9 & 3 & 1 & 6 & 7 & 5 & X & 2 & 8 \\
9 & 5 & 1 & 4 & 8 & 6 & 7 & 2 & X & 3 \\
5 & 1 & 6 & 2 & 7 & 9 & 4 & 8 & 3 & X
\end{array}\right)
$$ or
$$\left(\begin{array}{rrrrrrrrrr}
X & 7 & 2 & 6 & 3 & 1 & 8 & 4 & 9 & 5 \\
7 & X & 8 & 3 & 4 & 6 & 2 & 9 & 5 & 1 \\
2 & 8 & X & 5 & 9 & 4 & 7 & 3 & 1 & 6 \\
6 & 3 & 5 & X & 8 & 7 & 9 & 1 & 4 & 2 \\
3 & 4 & 9 & 8 & X & 5 & 1 & 6 & 2 & 7 \\
1 & 6 & 4 & 7 & 5 & X & 3 & 2 & 8 & 9 \\
8 & 2 & 7 & 9 & 1 & 3 & X & 5 & 6 & 4 \\
4 & 9 & 3 & 1 & 6 & 2 & 5 & X & 7 & 8 \\
9 & 5 & 1 & 4 & 2 & 8 & 6 & 7 & X & 3 \\
5 & 1 & 6 & 2 & 7 & 9 & 4 & 8 & 3 & X
\end{array}\right)$$ 

This answers your question about uniqueness. 
So it looks very plausible to me, that the completion can always be done for $n\geq 8$ and it is **not unique** for $n\geq 10$.


  [1]: https://i.sstatic.net/BZYli.png