Hadrons as tensors of flavor symmetry even though flavor symmetry is broken?

@anna's answer gives you what you really want to know in physics, but I will address some of your formal anxieties. A major theme is the sharp distinction between degeneracy symmetries (the Lie algebras of operators that commute or almost commute with the Hamiltonian) and spectrum-generating symmetries (the Lie algebras of operators that do not commute with the hamiltonian, and, in fact, move you from one rung of the spectrum to others).

For the quantum oscillator, the Heisenberg algebra $[a,a^dagger]=1$ does not commute with the number operator Hamiltonian: it ladders you up and down to non-degenerate states. For the Hydrogen atom, the spectrum-generating symmetries so(4,1) and so(4,2) connect states of different energy, as the Hamiltonian is not a function of their Casimir invariants, but, instead, contains "ladder" pieces moving eigenstates of it to different, non-degenerate, eigenstates. When one turns off such pieces, the SGA collapses to a basically dull degeneracy algebra.

Recall how su(3) works. On the one hand, in the limit of equal quark masses, it is a good degeneracy symmetry. But, we are far away from this limit. In fact, the strange quark mass differs from the u,d masses by more than $Lambda_{QCD}$, or the constituent quark mass, a third of the proton mass. The genius of flavor su(3) is first it tabulates all states made up of these quarks, a pretty tabulation. The su(4) pyramid does this as well.

But, importantly, second, it tells you how this symmetry is broken, by U-spin and V-spin operators, in a systematic, predictable, way: it is such amplitudes, couplings, Clebsches, etc... that expedite an awful lot of the heavy lifting involved in hadron interactions. (Doing this sort of thing with constituent quark wave functions is a godawful mess... you want to know how it's done, and it makes sense, but in all likelihood you won't use it in all but simple estimates, like magnetic moments.)

You may well do the same for flavor su(6), but our visual intuition is lacking in 5 dimensions, so I don't know anyone doing this. In a way, they do, when they segregate the 3 light quarks off the 3 heavy ones, in "WIsgur" stunts, and connect the "brown muck" QCD effects of each.

QCD is blind to all such structures: it couples the same way to all quarks, of whatever mass or flavor, but its effects vary with their masses. It doesn't not alter flavor.

As the other answer points out, such flavor groups are also broken by the EW symmetry, which does alter flavor, adding another layer of systematic complication to the picture.

It is fair to say that "symmetry operators" is an imperfect physics synonym for "Lie algebra generators", whose currents are not always even close to being conserved, as you observe. Lie theory, however, is so powerful it helps a lot even when it appears lost.

Now flavor numbers. These are mere tags reminding you what quark you are discussing. They correspond to independent rephasing of each flavor quark separately, and their currents are conserved and do nothing. QCD, unlike the weak interactions, does not mutate flavor, same as electromagnetism (which can still tell the difference of their charges).

As a result, flavor charges, like, e.g., S, are strictly conserved, outside the realm of the weak interactions. They are not traceless su(3) generators, obviously, and same for su(2), su(4)... So nothing breaks them, and QCD treats them all equally. They are not part of your "broken flavor symmetry" canard...

• Bonus problem. Can you see how the su(3) currents for $lambda_3$ and also $lambda_8$ are conserved, after all?