My understanding of this question is really two different questions. Let me answer each of these in turn.
1) What is the relation between the CKM and PMNS matrices?
To see how this works consider the relevant quark interaction terms without any choice of basis,
\begin{equation}
- m _d \bar{d} d - m _u \bar{u} u - i W _\mu \bar{d} \gamma ^\mu P _L \bar{u}
\end{equation}
Here $ m _d $ and $ m _u $ are completely arbitrary $ 3 \times 3 $ matrices.
We can redefine the down type quarks such that $ m _d $ is diagonal, $ d \rightarrow U _d d $. This matrix can then be reabsorbed into $ u $ (by a choice of basis for $u$) keeping the charged current diagonal. However, after this second redefinition we can't redefine the up type quarks again since we lost that freedom.
Therefore to have mass eigenstates we must introduce a mixing matrix which we call the CKM (this is often referred to as a product of the transformations on the down-type and up-type quarks but this is a bit unnecessary since we can always redefine one of either the down-type or up-type quarks to be in the diagonal basis). The CKM appears in the charged current interaction,
\begin{equation}
W _\mu \bar{d} \gamma ^\mu P _L \bar{u} = W _\mu \bar{d}' \gamma ^\mu P _L V _{ CKM}\bar{u} '
\end{equation}
Then we define a quark to be the mass eigenstates. The "cost" of this is that then we have to deal with uncertainty about which particle is produced in the charged current interaction since now particles of different generations can interact with the charged current. Its important here to note that this would not have been true if we called our ``quarks'' the fields that had a diagonal charged current.
That being said lets contrast with the charge lepton sector. Here we have,
\begin{equation}
- m _\ell \bar{\ell } \ell - m _\nu \bar{\nu } \nu - i W _\mu \bar{\ell } \gamma ^\mu P _L \bar{\nu }
\end{equation}
If the neutrinos were massless ($ m _\nu = 0 $) then we can just redefine the charged lepton basis such that their mass matrix is diagonal and we don't introduce any mixings into the charged current. However, if neutrinos do get a small mass then we have a choice we can diagonalize the neutrino matrix or leave the charged current diagonal.
On the other hand, unlike for the quarks, the mass eigenstates of the neutrino are almost impossible to produce. We have very little control over the neutrinos and they are typically made in one of the interaction eigenstates (in the basis in which $ m _\nu $ is nondiagonal), due to some charged current interaction. Thus the neutrinos are going to oscillate between the different mass eigenstates due to the state being in a superposition of energy eigenstates. Since we can't produce these mass eigenstates it is more convenient to call our "neutrinos" the states which we produce and let them oscillate.
Finally note that we often do diagonalize the neutrino matrix and define the analogue to the CKM known as the PMNS matrix, however this is more of a convenient way to parametrize the neutrino mass matrix then anything else.
2) Do quarks experience particle oscillations?
In general whenever the interaction eigenstates are not equal to the mass eigenstates particles can experience oscillations. In practice whether or not these oscillations are observable will depend on the interactions of the outgoing particles. Quarks interact significantly with their environment making their oscillations not observable in a physical experiment. To see how this plays out consider some collider producing down-type quarks (this can be say from top decays). The outgoing states will take the form,
$$|\rm outgoing\rangle = \#_1 |d\rangle +\#_2 |s \rangle+\#_3 |b \rangle$$
with the different coefficients determined by the CKM angle. When acted on by the time evolution operator, this state will mix into the other interaction eigenstates and hence when $|\rm outgoing \rangle$ propagates, it oscillates.
However, once these states are produced they are quickly "measured" by the environment through the subsequent processes such as showering and hadronization. The timescale for hadronization is $\Lambda_{QCD}^{-1} $ or a length scale of about a femtometer. This is way shorter than where we could place our detectors to see such oscillations. Once hadronization takes place the states decohere and quantum effects are no longer observable. Hence the linear combination is destroyed well before these particles are allowed to reach our detectors.
