Another way to phrase the main obstacle in positive characteristic is the following:

Although we can formulate the Hodge standard conjecture purely cycle-theoretically, in practice the only way we get a grip on it is through a cohomological representation. For example, in characteristic $0$ we use Hodge cohomology to prove the positivity statement of the pairing on primitive cohomology.

However, in positive characteristic, there isn't even a cohomology theory taking values in a field that admits a notion of positivity! The only known cohomology theories are over fields like $\mathbb Q_\ell$ and $W(k)$, neither of which admits the structure of an ordered field (because they have too many roots of unity).

Thus, one either has to come up with a cycle-theoretic proof that doesn't make reference to any cohomology theory (as far as I can tell, this would be new even in characteristic $0$), or one has to come up with a Weil cohomology theory taking values in an ordered field (like $\mathbb Q$ or $\mathbb R$, although already over $\mathbb F_{p^2}$ cohomology theories with values in one of these two fields provably do not exist by a result of Serre).

**Edit:** As suggested by Will Sawin below (Serre's original argument is the case $K = \mathbb R$):

**Claim.** There is no Weil cohomology theory with values in any totally ordered field $K$.

Indeed, there exists a supersingular elliptic curve $E$ over $\mathbb F_{p^2}$ such that $D = \operatorname{End}(E) \otimes_\mathbb Z \mathbb Q$ is a quaternion division algebra. It is non-split at $\infty$ (e.g. because it is nontrivial but split at all $v \neq p, \infty$), so it is given by $D = (a,b)$ for $a, b \in \mathbb Q_{<0}$. Such a division algebra never splits over an ordered field (since $a$ and $b$ have to remain negative in $K$, so $b$ can never be a norm of $K(\sqrt{a})/K$). $\square$

(In fact, using the Brauer–Hasse–Noether sequence from class field theory and explicit computation of the local invariant maps, one sees that $D$ can be represented as
$$D = \left\{\begin{array}{ll}(-1,-1) & p = 2,\\(-1,-p) & p \equiv 3 \pmod{4},\\(-q,-p) & p \equiv 1 \pmod{4},\end{array}\right.$$
where in the third case, $q \equiv 3 \pmod{4}$ is a prime whose residue class mod $p$ is a non-square. However, we only needed the negativity of $a$ and $b$.)