ADDED: it all works. In Dickson's METN, we find, theorem 117 on page 164, that $w^2 + u^2 - 6 z^2$ is the only equivalence class of "determinant" $-6,$ attributed to Ross. Then on page 170, exercise 2, we see that $w^2 + u^2 - 6 z^2$ integrally represents all numbers other than $$ 4^k \left( 16 m + 6 \right) \; , \; \; 9^k \left( 9 m + 6 \right) \; . $$
In particular, $9-6 = 3,$ and all other primes $p \equiv 3 \pmod 8$ are so represented.
ORIGINAL:
This is provisional, I will think about it overnight. The form
$$ w^2 - 2 y^2 + 3 y^2 - 6 z^2 $$
is anisotropic in $\mathbb Q_3$ three-adics. It is elementary to show that the form (integrally) represents the product of any two integers it integrally represents, that it does represent $-1,$ and that it integrally represents all (positive) primes
$$ p \equiv 1,5,7 \pmod 8 $$
A computer search strongly suggests that it also represents all primes $p \equiv 3 \pmod 8.$ If that can be proved, we are done. Note tht the output below has $x=y,$ meaning that we are looking at an indefinite ternary $w^2 + v^2 - 6 z^2.$
Mon Oct 28 11:10:15 PDT 2019
prime p == 3 mod 8, and demanding x = y:
p w x = y z
3 0 3 3 1
11 1 4 4 1
19 0 5 5 1
43 0 7 7 1
59 1 8 8 1
67 0 11 11 3
83 4 11 11 3
107 1 16 16 5
131 4 11 11 1
139 0 17 17 5
163 0 13 13 1
179 4 13 13 1
211 0 19 19 5
227 4 19 19 5
251 1 16 16 1
283 0 17 17 1
307 0 19 19 3
331 0 25 25 7
347 1 20 20 3
379 0 23 23 5
419 5 20 20 1
443 7 20 20 1
467 11 20 20 3
491 1 28 28 7
499 0 35 35 11
523 0 23 23 1
547 0 29 29 7
563 4 29 29 7
571 0 25 25 3
587 1 40 40 13
619 0 25 25 1
643 0 37 37 11
659 4 37 37 11
683 4 31 31 7
691 0 29 29 5
739 0 35 35 9
787 0 29 29 3
811 0 31 31 5
827 4 31 31 5
859 0 47 47 15
883 0 37 37 9
907 0 31 31 3
947 4 35 35 7
971 1 32 32 3
1019 1 32 32 1
1051 0 49 49 15
1091 4 35 35 5
1123 0 43 43 11
1163 7 40 40 9
1171 0 35 35 3
1187 4 35 35 3
1259 7 44 44 11
1283 8 35 35 1
1291 0 55 55 17
1307 1 40 40 7
1427 8 37 37 1
1451 1 40 40 5
1459 0 53 53 15
1483 0 47 47 11
1499 4 47 47 11
1523 8 53 53 15
1531 0 41 41 5
1571 4 43 43 7
1579 1 48 48 11
1619 5 40 40 1
1627 0 41 41 3
1667 5 44 44 7
1699 0 43 43 5
1723 0 47 47 9
1747 4 45 45 7
1787 1 44 44 5
1811 4 43 43 3
1867 7 48 48 9
1907 5 44 44 3
1931 1 44 44 1
1979 1 52 52 11
1987 3 52 52 11
2003 5 52 52 11
2011 0 55 55 13
2027 4 55 55 13
2083 0 53 53 11
2099 4 53 53 11
2131 4 51 51 9
p w x = y z
When $K$ is a field not of characteristic $2$, being able to solve $x^2 + y^2 + z^2 = 0$ nontrivially in $K$ is equivalent to being able to solve $x^2 + y^2 = -1$ in $K$, so I will think of the problem from that point of view.
I have checked explicitly that this can be done when $K$ is a quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$, as you say you did, and it can also be done when $K = \mathbf Q_2(\sqrt{2})$.
I'll list all seven quadratic extensions of $\mathbf Q_2$ in the order you did:
$$
\mathbf Q_2(\sqrt{-1}), \
\mathbf Q_2(\sqrt{-3}), \
\mathbf Q_2(\sqrt{-5}), \
\mathbf Q_2(\sqrt{-2}), \
\mathbf Q_2(\sqrt{-6}), \
\mathbf Q_2(\sqrt{-10}), \
\mathbf Q_2(\sqrt{2}).
$$
In the four fields $\mathbf Q_2(\sqrt{-1})$,
$\mathbf Q_2(\sqrt{-5})$,
$\mathbf Q_2(\sqrt{-2})$, and
$\mathbf Q_2(\sqrt{-10}),$
it is very easy to see a solution to $x^2 + y^2 = -1$:
$$
0^2 + \sqrt{-1}^2, \
2^2 + \sqrt{-5}^2, \
1^2 + \sqrt{-2}^2, \
3^2 + \sqrt{-10}^2.
$$
In $\mathbf Q_2(\sqrt{-3})$ we have a cube root of unity
$\zeta_3 = (-1 + \sqrt{-3})/2$, for which $1 + \zeta_3 + \zeta_3^2 = 0$, so
$$
-1 = \zeta_3 + \zeta_3^2 = \zeta_3^4 + \zeta_3^2.
$$
In $\mathbf Q_2(\sqrt{-6})$, after playing around a bit I got
$$
\left(\frac{2+\sqrt{-6}}{2}\right)^2 +
\left(\frac{2-\sqrt{-6}}{2}\right)^2 = -1.
$$
That leaves us with $\mathbf Q_2(\sqrt{2})$, whose ring of integers is $\mathbf Z_2[\sqrt{2}]$. I am not going to record here all the numerical work I did (by hand), but just report the final result: we can write
$$
-1 = x^2 + (\sqrt{2} + 2)^2
$$
for some $x \in \mathbf Z_2[\sqrt{2}]$ where $x \equiv 1 \bmod \sqrt{2}^3$. This follows from Hensel's lemma on $\mathbf Q_2(\sqrt{2})$ for the polynomial $f(X) = X^2 + 1 + (\sqrt{2}+2)^2 = X^2 + 7 + 4\sqrt{2}$:
$$
f(1) = 8 + 4\sqrt{2} \Longrightarrow |f(1)|_2 = \frac{1}{4\sqrt{2}}
$$
and
$$
f'(1) = 2 \Longrightarrow |f'(1)|_2^2 = \frac{1}{4}.
$$
Since $|f(1)|_2 < |f'(1)|_2^2$, Hensel's lemma tells us there is a unique solution to $f(x) = 0$ where $x \in \mathbf Z_2[\sqrt{2}]$ and
$|x - 1|_2 < |f'(1)|_2 = 1/2$, or equivalently $x \equiv 1 \bmod (\sqrt{2}^3)$. A more precise form of Hensel's lemma tells us that $|x - 1|_2 = |f(1)/f'(1)|_2 = (1/4\sqrt{2})/(1/2) = 1/2\sqrt{2} = 1/\sqrt{2}^3$, so $x \not\equiv 1 \bmod \sqrt{2}^4$.
We can compute $x$ using the binomial series for the square root:
$$
x = (-7 - 4\sqrt{2})^{1/2} = (1 - 4\sqrt{2} - 8)^{1/2}
$$
so
$$
x = \sum_{n \geq 0} \binom{1/2}{n}(-4\sqrt{2}-8)^n = 1 + \sum_{n \geq 1} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n
$$
Since ${\rm ord}_2\binom{1/2}{n} = -2n + s_2(n)$, where $s_2(n)$ is the sum of the base $2$ digits of $n$, and ${\rm ord}_2((4\sqrt{2}+8)^n) = 5n/2$, the $n$th term of the series
has $2$-adic valuation $5n/2 - 2n + s_2(n) = n/2 + s_2(n)$, so its $\pi$-adic valuation ($\pi = \sqrt{2}$ over the $2$-adics) is twice that: $n + 2s_2(n)$. Since $n+2s_2(n) \geq 7$ for all $n \geq 1$ other than $n = 1$, $2$, and $4$,
$$
x \equiv 1 + \sum_{n = 1, 2, 4} \binom{1/2}{n}(-1)^n(4\sqrt{2}+8)^n \bmod \pi^7\mathbf Z_2[\pi].
$$
Working this out modulo $\pi^7$,
$$
x \equiv 1 + \pi^3 + \pi^5 + \pi^6 \bmod \pi^7\mathbf Z_2[\pi].
$$
In "normal" notation, with $\pi = \sqrt{2}$, this approximate root of $f(X)$ is $9+6\sqrt{2} \bmod 8\sqrt{2}\mathbf Z_2[\sqrt{2}]$. As a check,
$$
(9 + 6\sqrt{2})^2 + (7 + 4\sqrt{2}) = 160 + 112\sqrt{2} = 16\sqrt{2}(5\sqrt{2} + 7),
$$
which is a multiple of $\pi^7 = 8\sqrt{2}$.
In what way did your work suggest that there might not be a solution to $x^2 + y^2 = -1$ in $\mathbf Q_2(\sqrt{2})$?
There is a high-brow explanation for why we found a solution in the particular quadratic extension $\mathbf Q_2(\sqrt{2})$. Being able to solve $x^2 + y^2 = -1$ in a local field $F$ (not of characteristic $2$) is the same as saying the Hilbert symbol $(-1,-1)_F$ is $1$ rather than $-1$. Letting $F = \mathbf Q(\sqrt{2})_v$ be a completion of $\mathbf Q(\sqrt{2})$, Hilbert reciprocity says the number of $v$ such that $(-1,-1)_{\mathbf Q_2(\sqrt{2})_v}$ is $-1$ is even. When $v$ is a place of odd residue field characteristic, $\mathbf Q(\sqrt{2})_v$ is a local field that is an extension of $\mathbf Q_p$ for an odd prime $p$, so that Hilbert symbol is $1$ since we can already solve $x^2 + y^2 = -1$ in $\mathbf Q_p$ (and actually in $\mathbf Z_p$). That leaves us with just three remaining places $v$ on $\mathbf Q(\sqrt{2})$: the two real places (corresponding to the two different embeddings of $\mathbf Q(\sqrt{2})$ into $\mathbf R$) and the unique place over $2$, corresponding to the prime $(\sqrt{2})$ in $\mathbf Z[\sqrt{2}]$.
We obviously can't solve $x^2 + y^2 = -1$ in $\mathbf R$, so
the Hilbert symbol $(-1,-1)_{\mathbf Q(\sqrt{2})_v}$ is $-1$ when $v$ is a real place. That leaves us with just one more place, over $2$, and since the Hilbert symbol is $-1$ at an even number of places it must be $1$ at the place over $2$.
By similar reasoning, if $K$ is an arbitrary real quadratic field, we can describe the non-archimedean places $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:
(1) If $v$ lies over an odd prime then there is a solution in $K_v$ since there already is one in $\mathbf Q_p$ for all odd $p$, and $K_v$ contains some $\mathbf Q_p$.
(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is just one $2$-adic place on $K$. We can solve $x^2 + y^2 = -1$ in $K_v$ by Hilbert reciprocity because we can solve it in every other completion except the two real completions, and two is even but three is not.
(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.
If $K$ is an arbitrary imaginary quadratic field, we can also determine the non-archimedean $v$ for which $x^2 + y^2 = -1$ has a solution in $K_v$:
(1) If $v$ lies over an odd prime then there is a solution in $K_v$ for the same reason as in the real quadratic case above.
(2) If $v$ lies over $2$ and $2$ does not split completely in $K$, then there is exactly one $2$-adic place on $K$. There is also exactly one archimedean place, with completion $\mathbf C$. We can solve $x^2 + y^2 = -1$ in $\mathbf C$, so Hilbert reciprocity implies we can solve $x^2 + y^2 = -1$ in $K_v$. In other words, we can solve $x^2 + y^2 = -1$ in every completion of $K$, so by Hasse-Minkowski we can solve $x^2 + y^2 = -1$ in $K$.
(3) If $v$ lies over $2$ and $2$ splits completely in $K$ then we can't solve $x^2 + y^2 = -1$ in $K_v$ since $K_v \cong \mathbf Q_2$.
The second item of each list for real and imaginary quadratic fields explains why we found solutions in every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ without needing to work in the completion itself (i.e., a solution in $\mathbf Q(i)$, $\mathbf Q(\sqrt{-3})$, and so on): you wrote every quadratic extension of $\mathbf Q_2$ other than $\mathbf Q_2(\sqrt{2})$ as a $2$-adic completion of an imaginary quadratic field, and in the imaginary quadratic fields that you chose, you can check that $2$ does not split completely (sometimes $2$ ramifies, sometimes $2$ is inert, but $2$ never splits completely). In the case of $\mathbf Q_2(\sqrt{2})$, we can write it as $\mathbf Q_2(\sqrt{-14})$, a completion of $\mathbf Q(\sqrt{-14})$ at its unique $2$-adic place.
Then results above say we have to be to solve $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{-14})$, and in fact a solution is $x = (2+3\sqrt{-14})/10$ and $y = (6-\sqrt{-14})/10$.
More generally, if $\mathbf Q_2(\sqrt{2}) = \mathbf Q_2(\sqrt{d})$ for some integer $d$ then ${\rm ord}_2(d) = 1$, so $\mathbf Q(\sqrt{d})$ ramifies at $2$ and thus there is a unique $2$-adic place on $\mathbf Q(\sqrt{d})$, so for $d < 0$ we'd be in the same situation as $\mathbf Q(\sqrt{-14})$: there will be a solution to $x^2 + y^2 = -1$ in $\mathbf Q(\sqrt{d})$.
Best Answer
The Witt decomposition is very non-unique; there is no unique hyperbolic part unless your form is hyperbolic. To find a maximal hyperbolic subspace, you'll want to find non-radical isotropic vectors and start pairing them up. From here I will just say null to mean non-radical isotropic. Specifically:
This process will yield null vectors $\nu_1, \nu_2, \dotsc, \nu_w$ and $\eta_1, \eta_2, \dots, \eta_w$ with $w$ the Witt index and $B(\nu_i, \eta_i) \ne 0$ but $B(\nu_i, \eta_j) = 0$ for $i \ne j$. The spaces $W = \mathrm{span}\{\nu_1,\dotsc,\nu_w\}$ and $W' = \mathrm{span}\{\eta_1,\dotsc,\eta_w\}$ are complementary fully-isotropic subspaces and $W \oplus W'$ is a maximal hyperbolic subspace.
Alternatively, we could follow the following process:
This process is equivalent since we can take $\nu_1 = e_1 + f_1$ and $\eta_1 = e_1 - f_1$, or similarly $e_1 = \nu_1 + \eta_1$ and $f_1 = \nu_1 - \eta_1$. Both methods have tradeoffs: in the first method, finding null vectors is easy, but finding non-orthogonal null vectors may be difficult; in the second method, finding orthogonal vectors is easy, but finding orthogonal vectors with opposite magnitudes may be difficult if your field doesn't have square roots.
For your example quadratic form $$ q(x, y, z) = x^2 + y^2 - 2z^2,\quad B(x_1, y_1, z_1, x_2, y_2, z_2) = x_1x_2 + y_1y_2 - 2z_1z_2 $$ we see that $$ e_1 = (1, 1, 0),\quad f_1 = (0, 0, 1). $$ are orthogonal and have opposite magnitudes. Since you know the Witt index is 1 then you know we're done and that $\mathrm{span}\{e_1,f_1\}$ is a maximal hyperbolic subspace.
Here is a third method: if we find an anisotropic subspace $K$ with codimension $2w + r$ where $r$ is the radical dimension, then $K^\perp = H\oplus R$ where $H$ is hyperbolic and $R$ is the radical. This follows from the Witt decomposition theorem and because $K\cap K^\perp = \{0\}$ for such a subspace. For your example $2w + r = 2$, so every non-isotropic vector gives us a maximal hyperbolic subspace. So every $v = (x, y, z)$ with $$ q(v) = x^2 + y^2 - 2z^2 \ne 0 $$ gives $\{v\}^\perp$ as a maximal hyperbolic subspace.