(partial answer; too long to fit as a comment)
The paper and GUAVA use different definitions for GF(4) orthogonality. With the paper's definition, Hamming(4,GF(4)) is self orthogonal. This gives a $4 \times 85$ matrix over GF(4), $H$, which corresponds to a $[[85,81,3]]$ code; taking $H$ and $\omega H$ gives an $8 \times 85$ matrix $H'$ that corresponds to the $[[85,77,3]]$ code in the paper. Note that the GF(4) rank of $H'$ is still 4, but it defines a different code than $H$. I can get these matrices and verify the weight enumerator of $C=(85,2^4)$ and its dual $C^\perp$ (page 15). So the "answer" to your question is then : take this $8 \times 85$ matrix and delete certain 76 of the 85 columns to get an $8 \times 9$ matrix which defines the $[[9,1,3]]$ code. These 76 columns are the support of all codewords in $C^\perp$ of weight 76. I don't beleive that the problem of finding these columns is tractable for codes this size but I could be wrong.
[update]
Instead of calculation which 76 columns to delete I did a random search of which 9 to keep. This found solutions fairly quickly (which means these are not that rate). Here's one :
[[1,3,2,1,1,1,0,0,0],
[0,2,1,2,2,0,2,1,0],
[2,0,1,2,0,3,0,0,0],
[3,0,3,0,2,3,3,0,1],
[2,1,3,2,2,2,0,0,0],
[0,3,2,3,3,0,3,2,0],
[3,0,2,3,0,1,0,0,0],
[1,0,1,0,3,1,1,0,2]]
$[0,1,2,3] \leftrightarrow [0,1,\omega,\omega^2]$