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. Well, this seems to do it: the genus and the spinor genus coincide unless the discriminant is divisible by $64$ or by $p^3$ for odd prime $p.$ Furthermore, they still coincide unless the diagonalization in $\mathbb Q_p$ is $a x^2 + bp y^2 + c p^2 z^2,$ whereupon we see that half the values $\pmod p$ are not represented, all $n$ with Legendre $(n,p) = -(a,p).$ More messy for $2,$ generally need $\pmod 8.$