I need to change your parameters a bit, but it sounded like you would be flexible, so let me suggest the following idea.
The idea only works when the number of stimuli is a prime number $p$. So if you need to test exactly 12 stimuli, then this is useless, but may be you can leave one out, or add a placebo/null test to the mix, and use this idea with $p=11$ or $p=13$.
The scheme has $p(p-1)$ rows and $k$ columns, where $k$ is any number between $2$ and $p$ inclusive.
Instead of letters A,B,... I use numbers $0,1,\ldots,p-1$ as table entries. One of the standard constructions for Latin squares is the following: First pick a parameter $m$ that is an integer in the range $1\le m<p$. Then put on row #$i$ and column #$j$ the number that equals the remainder of $i+mj$ when divided by $p$. This gives us a $p\times p$ Latin square. Call it $L(m,p)$. In this Latin square all the pairs of consecutive entries on all rows differ by $m$. Therefore this square alone is the very opposite of balanced. Within this square a zero is always followed by an $m$, a one by $m+1$ et cetera. Note that we count cyclically $\pmod p$, so an entry $p-m$ is always followed by $(p-m)+m=p\equiv 0$, and so forth.
Here comes the trick. We build a larger table with $p(p-1)$ rows by putting all the Latin squares $L(1,p)$, $L(2,p)$, $\ldots$, $L(p-1,p)$ on top of each other. By using all the possible values of $m$ we get a balanced table in the end!
We can take the $k$ first columns of this large table, and be done with it. Each entry occurs on all the columns exactly $p-1$ times. If two distinct entries, say $a$ and $b$, differ by $m=b-a$, the pair $(a,b)$ appears in the rows of $L(m,p)$ exactly $k-1$ times - once per each pair of adjacent columns, and doesn't appear on any other rows. There are no repetitions of stimuli within rows, as the rows are parts of rows of a Latin square.
Why doesn't this work with $p=12$? IOW, why do I insist that $p$ must be a prime? The reason is that the formula $i+m*j \bmod p$ gives a Latin square only, when $m$ is coprime to $p$. For example, if $p=12$ and $m=6$, then the rows of $L(6,12)$ look like 0,6,0,6,...; 1,7,1,7,...
With $p=13$ you get 156 rows, so the table may be too large for you. Another possibly troubling feature of this construction is the following. The rows of the Latin square $L(1,p)$ look like 0,1,2,...; 1,2,3,...; so they have intersections of $k-1$ elements. This may be bad for eliminating secondary correlations from your test. If you do $k=6$ tests per participant, then five participants will see stimulus #1 followed by stimulus #2. That's ok, but I am a bit troubled by the fact that four out of those five will see stimulus #3 next. And the fifth person won't see anything, because his/her day ends after stimulus #2. Similar patterns appear in other component squares $L(m,p)$. If these shortcomings make the construction unusable, then I apologize for wasting your time.
[Edit: Oh boy, I should learn to proofread and not post in such haste. I apologize for the mostly illegible first version :-(]
Best Answer
What you have aren't balanced latin squares mainly because they aren't even squares.:) You need to have the same number of rows as columns. So a balanced latin square of order 6 would be
$$\begin{matrix} 1&2&3&6&4&5\\ 2&3&4&1&5&6\\ 3&4&5&2&6&1\\ 4&5&6&3&1&2\\ 5&6&1&4&2&3\\ 6&1&2&5&3&4\\ \end{matrix} $$
In other words you need to have the same number of participants as condition to be able use the balanced latin square approach. Or rather you need to have the same number of participant groups. So really the number of participants must be a multiple of the number of conditions/treatments what have you.
Edit A nice brief exposition can be found here. It also includes an explanation of how to deal with the situation that the number of conditions you have is odd.