quantum field theory - Why do we antisymmetrize the baryon's wavefunction but not the meson's?

A many-body wavefunction of identical fermions must be antisymmetrized because of fermionic statistics. We don’t antisymmetrize the meson wavefunction because it contains a quark and an antiquark, and they are not identical fermions. But in particle physics, we do antisymmetrize the baryon wavefunction even though it may consist of 3 non-identical quarks of different flavors.

What is the fundamental reason for this? Why is the baryon wavefunction anti-symmetrized? The different colors of quarks have exact SU(3) color symmetry, so it is probably okay to regard them as identical fermions [And we write it as a color singlet, fully antisymmetric, see below.] But the different flavors of quarks have no exact SU($N_f$) flavor symmetry, so why are different flavors of quarks regarded as identical?

There is some discussion in Griffiths Introduction to Elementary Particles p.184,

The wave function of a baryon consists of several pieces; there is the spatial part, describing the locations of the three quarks; there is the spin part, representing their spins; there is a flavor component, indicating what combination of u, d, and s is involved; and there is a color term, specifying the colors of the quarks: (space) (spin) (flavor) (color). It is the whole works that must be antisymmetric under the interchange of any two quarks.

Notice that a subtle extension of the notion of “identical particle” has implicitly been made here, for we are treating all quarks, regardless of color or even flavor, as different states of a single particle.

Unfortunately, Griffiths does not give any further reasoning.

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Notes on the question:

1. 
   

   I am not asking why the baryon color wavefunction is a singlet. I already knew it had to be $(| rgb \rangle -| rbg \rangle + | gbr \rangle - | grb \rangle + | brg \rangle -| bgr \rangle )/\sqrt{6}$.

   
   


2. 
   

   Explicitly, consider the baryon octet $\Delta^+$ (from the $10$ in $3 \otimes 3 \otimes 3= 10 \oplus 8 \oplus 8 \oplus 1$) with a total spin $3/2$ and spin $z$ as $-1/2$, with wavefunction: $$\begin{aligned} &(| rgb \rangle -| rbg \rangle + | gbr \rangle - | grb \rangle + | brg \rangle -| bgr \rangle )/(\sqrt{6})\\ \otimes&(uud+udu+duu)/(\sqrt{3})\\ \otimes&(\downarrow \downarrow \uparrow +\downarrow \uparrow \downarrow +\uparrow \downarrow \downarrow )/(\sqrt{3}) \end{aligned}$$ The answer should argue why this wavefunction needs to be antisymmetrized, though $u$ and $d$ are non-identical.

   
   


3. 
   
   

   My question is related to this nice Phys.SE post (which I happily vote it up +1), but the accepted answer does not explain anything new, only re-iterate the known fact.

   
   


4. 
   

   Related older Reference: Phys. Rev., Vol. 50, 846 1936, On Nuclear Forces by B. CASSEN AND E. U. CONDON