THIRD answer: It turns out the condition I gave in my second answer is necessary and sufficient for existence; I can also show a two parameter family, I'm afraid to minimize could be algorithmic but not formulaic. i will have time for that aspect later. Given Gram matrix with $A,B,C$ as before,
we have the equation
$$ A \alpha \beta + B (\alpha \delta + \beta \gamma) + C \gamma \delta. $$
So need to have $iA + jB + kC = 0,$ with $j^2 - 4ik = w^2.$
So, take integer parameters $s,t,$ then
$$ \alpha = 2i s, \; \; \beta = 2 it, \; \; \gamma = (j-w)s, \; \; \delta = (j+w)t. $$ With these values, the vector $( \alpha \beta, \alpha \delta + \beta \gamma, \gamma \delta)$ is a scalar (rational) multiple of $(i,j,k),$ and we have constructed orthogonal vectors in the lattice. It is possible that $\gcd(\alpha, \beta, \gamma,\delta)> 1$ for some values of $(s,t)$ but not others.
Furthermore, minimization of the determinant $|\alpha \delta - \beta \gamma|$ needs work, although it is a multiple of $w$ by construction. Here is a start:
$$ \alpha \delta - \beta \gamma = 4iwst. $$ Need to think about what that means, with varying GCD's and the possibility of zero values for $s,t.$ SIGH. Added: no, if one of $s,t$ is zero, one vector in the orthogonal pair is just the zero vector, so we may rule out that possibility. Good. So, it may not be smallest, but $s=t=1$ gives an orthogonal pair.
$$
\left(
\begin{array}{rr}
2i & j-w \\
2i & j+w
\end{array}
\right)
\left(
\begin{array}{rr}
A & B \\
B & C
\end{array}
\right)
\left(
\begin{array}{rr}
2i & 2i \\
j-w & j+w
\end{array}
\right) =
\left(
\begin{array}{cc}
4i^2 A + 4i(j-w)B + (j-w)^2C & 0 \\
0 & 4i^2 A + 4i(j+w)B + (j+w)^2C
\end{array}
\right)
$$
If $i=0,$ thus $jB + kC=0,$ we get
$$
\left(
\begin{array}{rr}
j & k \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
A & B \\
B & C
\end{array}
\right)
\left(
\begin{array}{rr}
j & 0 \\
k & 1
\end{array}
\right) =
\left(
\begin{array}{cc}
j^2 A + 2jkB + k^2C & 0 \\
0 & C
\end{array}
\right)
$$
When $j=0,$ the combination of $\gcd(i,j,k)=1$ and $j^2 - 4 i k = w^2$ allows us to demand, in integers,
$$ i = x^2, \; \; \; k = -y^2, \; \; \; w = 2 x y, $$
with $x^2 A - y^2 C = 0.$ Then
$$
\left(
\begin{array}{rr}
x & -y \\
x & y
\end{array}
\right)
\left(
\begin{array}{rr}
A & B \\
B & C
\end{array}
\right)
\left(
\begin{array}{rr}
x & x \\
-y & y
\end{array}
\right) =
\left(
\begin{array}{cc}
A x^2 - 2 B x y + C y^2 & 0 \\
0 & A x^2 + 2 B x y + C y^2
\end{array}
\right)
$$
Anyway, these show existence for any explicit $(i,j,k)$ triple of integers such that $iA + j B + k C = 0.$ These also give upper bounds on the determinants of the change of basis matrices, again $|\alpha \delta - \beta \gamma|.$ In the case that the lattice cannot be scaled to an integral lattice, I am not confident about giving an explicit recipe for the smallest determinant that works; I suggest using these as upper bounds for a computer search.