Yes, it's a normal field theory, so you may derive the equations of motion. They will be the ordinary Maxwell's equations for the electromagnetic field
$$ \partial_\nu F^{\mu\nu} = j^\nu $$
with $j^\mu$ calculated as the sum of the conserved currents for the Dirac field and for the Higgs fields, combined with the Dirac equation coupled to the electromagnetic field (with some Yukawa interaction $y\cdot \phi\psi$ terms), and the Klein-Gordon equation for a charged scalar field with some $V'(\phi)$ and $\psi \psi$ terms added in the right hand side etc.
In fact, the Yukawa Lagrangian is (more or less) only the term $\mathcal{L}_Y = -g \bar{\psi}\psi \phi$. The (massless) Dirac Lagrangian for fermions and Klein-Gordon Lagrangian (plus potential) for the Higgs are not shown in your formula. The main difference between the Yukawa Lagrangian and the simpler $-g \bar{\psi}\psi \phi$ is that the Standard Model has several fermionic fields, which are coupled in this Lagrangian.
The $h$ matrices are called Yukawa couplings, and are the equivalent to $g$ above. They are matrices instead of numbers because you will end up with terms like $-g \bar{s}d \phi$ (more to that later).
To understand the difference between $\bar{q}_{aL}$ and $\bar{u}_{aL}$ you have to remember that left-handed fermions come in doublets of isospin. The Higgs field is also a doublet of isospin $$\phi =\frac{1}{\sqrt{2}} \begin{pmatrix}\phi_1 + i \phi_2\\ \phi_0 +i\phi_3 \end{pmatrix}\ . $$ After electroweak spontaneous symmetry breaking, only the $\phi_0$ components has a non-zero value in the vacuum. If we replace in the Yukawa Lagrangian for $d$-type quarks or leptons, we get
$$-\mathcal{L}_d = \sum_{a,b} h_{ab}^{(d)} \begin{pmatrix} \bar{u}_{aL} & \bar{d}_{aL} \end{pmatrix}\begin{pmatrix} 0 \\ \phi_0 \end{pmatrix} d_{bR} = \sum_{a,b} h_{ab}^{(d)} \phi_0 \bar{d}_{aL} d_{bR}\ . $$
So the fact that the Higgs field breaks the isospin symmetry selects only the lower component ($T_3 = +\frac{1}{2}$) of the isospin doublet. This explains the mass of $d$-type quarks and charged leptons.
But you need also a mass term for the $u$-type quarks. This is the purpose of the first term in the Yukawa Lagrangian, wich is a sort of "transposed version" of the other two terms. Here you need the charge-conjugated Higgs field $\hat{\phi}$. This Lagrangian reads
$$-\mathcal{L}_u = \sum_{a,b} h_{ab}^{(u)} \begin{pmatrix} \bar{u}_{aL} & \bar{d}_{aL} \end{pmatrix}\begin{pmatrix} \phi_0 \\ 0 \end{pmatrix} u_{bR} = \sum_{a,b} h_{ab}^{(u)} \phi_0 \bar{u}_{aL} u_{bR}\ .$$
To understand the reason of the (non)-diagonal matrices, note that the Lagrangian is invariant under unitary transformations $$u_R^i \to V^{ij}_{uR} u_R^j \qquad \qquad d_R^i \to V^{ij}_{dR} d_R^j\\ u_L^i \to V^{ij}_{uL} u_L^j \qquad \qquad d_L^i \to V^{ij}_{dL} d_L^j$$
you can choose these matrices in order to diagonalize the masses $$D^{u,d} = \frac{v}{\sqrt{2}} V_{u,d L} h^{(u,d)} V_{u,d R}^\dagger\ . $$
But four different matrices is just overkill. You can't diagonalize $h^{(u)}$ and $h^{(d)}$ simultaneously, so it is chosen to diagonalize $u$-type quarks and rotate the $d$-type quarks using the CKM matrix. This election is just a convention, it could be done the other way.
The physical significance of this quark mixing is that charged can connect, with a small probability, quarks of different generations. You can describe it saying that the weak $d'$ quark is a superposition of $d$ and $s$ quark mass eigenstates, or saying that the weak $u'$ quark is a superposition of the $u$ and $c$ quark mass eigenstates; it makes no difference.
Best Answer
Because the "theory" you write down doesn't exist. It's just a logically incoherent mixture of apples and oranges, using a well-known metaphor.
One can't construct a theory by simply throwing random pieces of Lagrangians taken from different theories as if we were throwing different things to the trash bin.
For numerous reasons, loop quantum gravity has problems with consistency (and ability to produce any large, nearly smooth space at all), but even if it implied the semi-realistic picture of gravity we hear in the most favorable appraisals by its champions, it has many properties that make it incompatible with the Standard Model, for example its Lorentz symmetry violation. This is a serious problem because the terms of the Standard Model are those terms that are renormalizable, Lorentz-invariant, and gauge-invariant. The Lorentz breaking imposed upon us by loop quantum gravity would force us to relax the requirement of the Lorentz invariance for the Standard Model terms as well, so we would have to deal with a much broader theory containing many other terms, not just the Lorentz-invariant ones, and it would simply not be the Standard Model anymore (and if would be infinitely underdetermined, too).
And even if these incompatible properties weren't there, adding up several disconnected Lagrangians just isn't a unified theory of anything.
Two paragraphs above, the incompatibility was presented from the Standard Model's viewpoint – the addition of the dynamical geometry described by loop quantum gravity destroys some important properties of the quantum field theory which prevents us from constructing it. But we may also describe the incompatibility from the – far less reliable – viewpoint of loop quantum gravity. In loop quantum gravity, one describes the spacetime geometry in terms of some other variables you wrote down and one may derive that the areas etc. are effectively quantized so the space – geometrical quantities describing it – are "localized" in some regions of the space (the spin network, spin foam, etc.). This really means that the metric tensor that is needed to write the kinetic and other terms in the Standard Model is singular almost everywhere and can't be differentiated. The Standard Model does depend on the continuous character of the spacetime which loop quantum gravity claims to be violated in Nature. So even if we're neutral about the question whether the space is continuous to allow us to talk about all the derivatives etc., it's true that the two frameworks require contradictory answers to this question.