The valence quark content of a baryonic state is $qqq$, that is to say three identical quarks make up the content of the bound state amongst a sea of partons. Each quark therein carries a series of quantum numbers which serve to denote the different possible states of the quark and, collectively, must be such that a simultaneous permutations of the degrees of freedom yields an overall antisymmetric state, in accordance with the fact the quantum state must obey Fermi-Dirac statistics.
For non excited states (the so called lowest lying states), the spatial component of the direct product is always symmetric ( $S$-wave orbital angular momentum) and colour is antisymmetric so the combination $|\text{spin} \rangle \otimes |\text{flavour} \rangle$ must be symmetric. The observable states transform under irreducible multiplets of (approximate) $SU(3)$ flavour symmetry and so the '$|uds \rangle$' state is actually a flavour symmetric combination in the $10$ decuplet, a flavour antisymmetric combination in the $1$, and appears with mixed flavour symmetries in the remaining two octets in agreement with the group theoretic decomposition $3 \otimes 3 \otimes 3 = 1 \oplus 8 \oplus 8 \oplus 10$.
So. e.g for the totally flavour symmetric combination, the spin state of the wavefunction must be in one of the symmetric spin $3/2$ states.
The extension that Griffiths is proposing isn't as drastic as you think. When you first learn about identical particles, you antisymmetrize the spatial wavefunction alone. Later, you add the spin and antisymmetrize the combined spin and spatial wavefunctions. But this is a major conceptual leap, because a spin up and spin down electron are not identical particles. The Pauli exclusion principle doesn't apply to them; you can put a spin up and spin down electron in the same state.
Now the jump to including the 'flavor wavefunction' is the same. An up quark and a down quark are not identical particles, and they can indeed be put in the same state. You might argue that including spin is different because you can flip it with a rotation, but the up and down quark states can be flipped by an isospin rotation; the situation is really exactly analogous.
So spin and flavor are on the same grounds; neither of them are correctly justified with the standard 'quantum mechanics' argument of interchanging the positions of two particles, since that only applies to the spatial wavefunction. Instead, we have to go to quantum field theory. The reason the spatial wavefunction is antisymmetric is because the creation operators anticommute,
$$a_x^\dagger a_y^\dagger |0 \rangle = - a_y^\dagger a_x^\dagger |0 \rangle.$$
But the spatial indices $x$ and $y$ are just like any other kind of index; in reality we should have indices for flavor, color, and spin as well. Then by the same logic, the overall antisymmetry of the flavor/color/spin/space wavefunction follows from the fact that any fermionic creation operators anticommute.
So, why we don't antisymmetrize the meson wavefunctions? Consider a box of 10 electrons and an 11th electron a light year away. The wavefunction for all 11 electrons must be antisymmetrized by the general argument above, but nothing changes if we only antisymmetrize the first 10. Concretely, that's because the only observable thing the antisymmetrization does is add an exchange force, and the 11th electron will never be close enough to feel it. Conceptually, we can say the 11th electron is distinguishable by virtue of its position -- it's 'the one that's far away'.
Similarly, if we have an atom with 10 spin up electrons and 1 spin down electron, we can treat the spin down electron as separate from the antisymmetrized wavefunction of the other 10, distinguishable by its spin. The true wavefunction of all 11 is still fully antisymmetric by the argument given above, but nothing goes wrong here if we forget that; we won't run afoul of the Pauli exclusion principle by accident.
Now we turn to mesons and baryons.
- In the case of mesons, the quarks are always effectively distinguishable because only one of them is an antiparticle, or equivalently, only one has anticolor.
- For baryons with all three quarks distinct, the same reasoning holds; the particles are effectively distinguishable by their flavors. For example, consider $uds$ baryons. If we treat the particles as distinguishable, there are $2^3 = 8$ spin assignments, and accordingly there are $8$ low-energy $uds$ baryons. (They're the four spin states of the ${\Sigma^*}^0$ and the two spin states of the $\Sigma^0$ and $\Lambda$.)
- For baryons with some quarks the same, none of the quarks are effectively distinguishable; we must account for the full antisymmetry. For example, there are only $4$ baryons with quark content $uuu$, not $8$. (They're the four spin states of the $\Delta^{++}$.) Similarly there are only $6$ with quark content $uud$. (They're the four spin states of the $\Delta^+$ and the two of the proton.)
So the particles in a meson are always effectively distinguishable, but the particles in a baryon are only sometimes. It's more economical to treat all the baryons the same way, so that's where the rule to 'antisymmetrize baryons but not mesons' comes from.
Best Answer
These two Figures are excerpt of a page in : A Modern Introduction to Particle Physics, by Fayyazuddin & Riazuddin, 2nd Edition 2000.
The first shows the well-known octet of the mixed antisymmetric tensor $\:\boldsymbol{8}\:$ while the second shows the octet of the mixed symmetric tensor $\:\boldsymbol{8'}\:$. I don't know what particles, if any, are represented by the latter octet.
See also my answer here : Symmetry in terms of matrices. Therein octet $\:\boldsymbol{8}\:$ is produced by the mixed antisymmetric tensor $\:Y_{ijk}\:$, see equations (B.25) and (B.35), while octet $\:\boldsymbol{8'}\:$ is produced by the mixed symmetric tensor $\:X_{ijk}\:$, see equations (B.24) and (B.37).
The Figures below are excerpts from : A Modern Introduction to Particle Physics-Volume 1: Quantum Field Theory and Particles, by Y.Nagashima, Edition 2010.
In 'QUARKS AND LEPTONS: An Introductory Course in Modern Particle Physics', F.Halzen-A.Martin, Edition 1984, we meet the following concerning the spin-up proton: We take the mixed-antisymmetric and mixed-symmetric states
\begin{align} \mathrm p_{_A} & =\sqrt{\tfrac12}\left(\mathrm u \mathrm d- \mathrm d\mathrm u \right)\mathrm u \:\: \left\{\in \boldsymbol{8}_{_{MA}}\equiv \boldsymbol{8}\right\} \tag{2.60}\\ \mathrm p_{_S} & =\sqrt{\tfrac16}\bigl[\left(\mathrm u \mathrm d+ \mathrm d\mathrm u \right)\mathrm u-2\mathrm u \mathrm u \mathrm d \bigr]\:\: \left\{\in \boldsymbol{8}_{_{MS}}\equiv \boldsymbol{8'}\right\} \tag{2.62} \end{align} The state $\:\mathrm p_{_A} \:$ is the first member of octet $\:\boldsymbol{8}\:$ shown in the first Figure above while the state $\:\mathrm p_{_S} \:$ is the first member of octet $\:\boldsymbol{8'}\:$ shown in the second Figure.
These are multiplets produced in $\:\rm SU(2)-$isospin and by analogy the $\:\rm SU(2)-$spin antisymmetric, symmetric multiplets are produced by replacing in (2.60), (2.62)
\begin{align} \mathrm u & \quad \Longrightarrow \quad \uparrow \nonumber\\ \mathrm d & \quad \Longrightarrow \quad \downarrow \tag{01} \end{align} so \begin{align} \chi\left(M_A\right) & =\sqrt{\tfrac12}\left(\uparrow \downarrow\uparrow -\downarrow\uparrow\uparrow \right) \nonumber\\ \chi\left(M_S\right) & =\sqrt{\tfrac16}\left(\uparrow \downarrow\uparrow +\downarrow\uparrow\uparrow -2\uparrow\uparrow \downarrow \right) \tag{2.65} \end{align} Next the spin-up proton is derived from \begin{equation} \vert \mathrm p\!\uparrow\, \rangle=\sqrt{\tfrac12}\bigl[\mathrm p_{_A}\chi\left(M_A\right)+\mathrm p_{_S}\chi\left(M_S\right) \bigr] \tag{02} \end{equation} where \begin{align} \mathrm p_{_A}\chi\left(M_A\right) & \simeq \left(\mathrm u \mathrm d \mathrm u-\mathrm d \mathrm u \mathrm u\right)\left(\uparrow \downarrow\uparrow -\downarrow\uparrow\uparrow \right) \nonumber\\ & =\bigl(\mathrm u\!\!\uparrow\!\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow -\mathrm u\!\!\downarrow\!\mathrm d\!\!\uparrow\!\mathrm u\!\!\uparrow -\mathrm d\!\!\uparrow\!\mathrm u\!\!\downarrow\!\mathrm u\!\!\uparrow +\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow\!\mathrm u\!\!\uparrow \bigr) \tag{03} \end{align} and \begin{align} \mathrm p_{_S}\chi\left(M_S\right) & \simeq\left(\mathrm u \mathrm d\mathrm u + \mathrm d\mathrm u \mathrm u-2\mathrm u \mathrm u \mathrm d \right)\left(\uparrow \downarrow\uparrow +\downarrow\uparrow\uparrow -2\uparrow\uparrow \downarrow \right) \nonumber\\ & =\mathrm u\!\!\uparrow\!\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow + \mathrm u\!\!\downarrow\!\mathrm d\!\!\uparrow\!\mathrm u\!\!\uparrow-2 \mathrm u\!\!\uparrow\!\mathrm d\!\!\uparrow\!\mathrm u\!\!\downarrow +\mathrm d\!\!\uparrow\!\mathrm u\!\!\downarrow\!\mathrm u\!\!\uparrow +\mathrm d\!\!\downarrow\!\mathrm u\!\!\uparrow\!\mathrm u\!\!\uparrow +\cdots \tag{04} \end{align} and finally
\begin{align} \vert \mathrm p\!\uparrow\, \rangle & =\sqrt{\tfrac{1}{18}}\bigl[\mathrm u \mathrm u \mathrm d \left(\uparrow \downarrow\uparrow +\downarrow\uparrow\uparrow -2\uparrow\uparrow \downarrow \right)+\mathrm u \mathrm d \mathrm u \left(\uparrow\uparrow\downarrow +\downarrow\uparrow\uparrow -2\uparrow\downarrow \uparrow \right)+\mathrm d \mathrm u \mathrm u \left(\uparrow\downarrow \uparrow+\uparrow\uparrow\downarrow -2\downarrow \uparrow\uparrow \right)\bigr] \nonumber\\ &=\sqrt{\tfrac{1}{18}}\bigl[\mathrm u\!\!\uparrow\!\mathrm u\!\!\downarrow\!\mathrm d\!\!\uparrow +\mathrm u\!\!\downarrow\!\mathrm u\!\!\uparrow\!\mathrm d\!\!\uparrow -2\mathrm u\!\!\uparrow\!\mathrm u\!\!\uparrow\!\mathrm d\!\!\downarrow +\text{permutations}\bigr] \tag{2.71} \end{align}
From above notes we conclude that the real baryon $\:(1/2)^{+}\:$ octet is a combination of the two octets with mixed symmetry $\:\boldsymbol{8},\boldsymbol{8'}$.