Does every $9 \times 9$ Latin square contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$? 
Q: Does every $9 \times 9$ Latin square on the symbol set $\{1,2,\ldots,9\}$ contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?

This one has $1728$ such submatrices, which is as low as I've gotten:
$$
\begin{bmatrix}
6 & 7 & 8 & 9 & 1 & 4 & 2 & 3 & 5 \\
5 & 6 & 1 & 7 & 2 & 8 & 3 & 4 & 9 \\
9 & 1 & 6 & 2 & 4 & 3 & 7 & 5 & 8 \\
4 & 5 & 3 & 6 & 8 & 7 & 1 & 9 & 2 \\
1 & 2 & 4 & 8 & 3 & 5 & 9 & 6 & 7 \\
2 & 3 & 7 & 4 & 9 & 6 & 5 & 8 & 1 \\
8 & 9 & 2 & 3 & 5 & 1 & 6 & 7 & 4 \\
7 & 8 & 5 & 1 & 6 & 9 & 4 & 2 & 3 \\
3 & 4 & 9 & 5 & 7 & 2 & 8 & 1 & 6 \\
\end{bmatrix}$$
It doesn't seem likely that random Latin squares will help much; they average in the thousands of such submatrices.  The one above is the best random Latin square I've found so far (although, I haven't busted a gut doing this; it seems like it won't work anyway).
The groups of order $9$ have lots ($C_9$ has $5832$ and $C_3 \times C_3$ has $19440$).

This question was motivated by answering this math.SE question which asks if any $9 \times 9$ Latin square can have its rows and columns permuted to give a sudoku square.
One way to find an explicit counterexample would be to find a $9 \times 9$ Latin square with no $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$.  But this attempt didn't work since I couldn't find one.  Hence my question.

How about this one?
$$
\begin{bmatrix}
  1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
  2 & 1 & 4 & 3 & 6 & 7 & 8 & 9 & 5 \\
  3 & 4 & 2 & 1 & 8 & 9 & 5 & 6 & 7 \\
  4 & 3 & 1 & 2 & 7 & 8 & 9 & 5 & 6 \\
  5 & 6 & 9 & 8 & 2 & 3 & 4 & 7 & 1 \\
  6 & 7 & 5 & 9 & 1 & 2 & 3 & 4 & 8 \\
  7 & 8 & 6 & 5 & 9 & 1 & 2 & 3 & 4 \\
  8 & 9 & 7 & 6 & 4 & 5 & 1 & 2 & 3 \\
  9 & 5 & 8 & 7 & 3 & 4 & 6 & 1 & 2 \\
\end{bmatrix}$$
I discovered it by modifying a Latin Square solver to add the restriction that no 3x3 submatrix can contain 9 different values. It found this in about 1 hour.
