Regarding the question in "Motivation": a $1$-factorization of $K_{n,n}$ is essentially an $n\times n$ Latin square, and a $3$-chromatic $K_{3,3}$ would be a proper Latin subsqare.

This paper shows that there are Latin $n\times n$ squares with no proper Latin subsquares for $n=pq \neq 6$ where $p,q$ are distinct primes. They claim the for $n$ prime it is well known that the multiplication table of the group of order $p$ is a Latin square with no proper Latin subsquares. Indeed, a proper Latin subsquare would correspond to a nontrivial proper subgroup.

Constructions of Latin $n\times n$ squares with no proper Latin subsquares are known for all odd $n$: http://www.sciencedirect.com/science/article/pii/S0195669805000909

Regarding the question in "Motivation": a $1$-factorization of $K_{n,n}$ is essentially an $n\times n$ Latin square, and a $3$-chromatic $K_{3,3}$ would be a proper Latin subsqare.

This paper shows that there are Latin $n\times n$ squares with no proper Latin subsquares for $n=pq \neq 6$ where $p,q$ are distinct primes. The author claims that for $n$ prime it is well known that the multiplication table of the group of order $p$ is a Latin square with no proper Latin subsquares. Indeed, a proper Latin subsquare would correspond to a nontrivial proper subgroup.

Constructions of Latin $n\times n$ squares with no proper Latin subsquares are known for all odd $n$: http://www.sciencedirect.com/science/article/pii/S0195669805000909