Two parts. If the form is anisotropic, there is a specific prime $p$ for which the form is anisotropic. The problem is that the form does not integrally represent anything in the $p$-adic squareclass of the discriminant of the form. This follows from just a few pages in Cassels, rational Quadratic Forms, which I recommend; mostly pages 58-59. 

Next, it is possible to have exceptions that do not directly arise from congruences. The famous 1951 example of Siegel is a genus with two forms,
$$  x^2 - 2 y^2 + 64 z^2, $$
$$  (2x+z)^2 - 2 y^2 + 16 z^2.  $$
Note that the binary $4x^2 + 4 xz + 17 z^2$ is indeed in the same genus as $x^2 + 64 y^2.$ However, while $x^2 + 64 y^2$ is a fourth power in the class group, $4x^2 + 4 xz + 17 z^2$ is a square but not a fourth power.
Now, $(2x+z)^2 - 2 y^2 + 16 z^2$ fails to represent any $m^2,$ where all prime factors of $m$ are $\pm 1 \pmod 8.$ Now, with $  x^2 - 2 y^2 + 64 z^2, $ we can take $x=y=8, z=1$ to get $0.$ Isotropic. At the same time, neither form represents any $\pm 3 \pmod 8.$

A more recent example of Schulze-Pillot and Xu is
$$  x^2 + 100 y^2 - 5 z^2, $$
$$  4 x^2 + 25 y^2 - 5 z^2.  $$
Note that the binary $4x^2  + 25 y^2$ is  in the same genus as $x^2 + 100 y^2.$ However, while $x^2 + 100 y^2$ is a fourth power in the class group, $4x^2 +  25 y^2$ is a square but not a fourth power.
Now, $4 x^2 + 25 y^2 - 5 z^2$ fails to represent any $m^2,$ where all prime factors of $m$ are $\pm 1 \pmod 5.$ Now, with $  x^2 + 100 y^2 - 5 z^2, $ we can take $x=5, y=1, z=5$ to get $0.$ Isotropic. At the same time, neither form represents any $\pm 3 \pmod 5.$

For the moment, I think that we cannot have spinor exceptional integers (as in the two examples) without congruence obstructions, even when all is isotropic. I do not have an immediate proof.