If Higgs field is present everywhere in the universe, then why can't we use it to find the absolute motion of an object relative to it? Why can't we describe absolute motion relative to Higgs field?
 A: The Higgs field is a quantum field, like all the other elementary particle fields.
It is true that the quantum fields fill all of spacetime but they are not physical objects (unlike the hypothetical aether through which light was originally believed to propagate) and you cannot measure your speed relative to them.
In fact quantum field theory is specifically formulated to be Lorentz covariant so the quantum fields behave the same way for all observers. This means there is no absolute frame of reference, so there is no such thing as an absolute speed.
A: The Higgs boson has been observed, and this observation is the capstone on the standard model of particle physics. The model is validated, and the Higgs field is part of the mathematical construct that  fits  the grand majority of particle data up to the present.
Note the word model. It means a self consistent mathematical theory with extra axioms to correlate the mathematics to the data. This model by construction is Lorenz covariant, because data in particle physics up to now validate fully Lorentz covariance. 
Lorenz transformations are relative , there is no absolute frame of reference possible, and the same is true for the Higgs field, which is a scalar.
A: Because a scalar field constant across spacetime is in a Lorentz invariant state.
A: Higgs field before symmetry breaking is an $U(2)$ valued scalar field, associating to each point in spacetime an $U(2)$ matrix.
The fact that it is scalar means that it does not change under Lorentz boosts and rotations, so you cannot distinguish frames by the value of the Higgs field, that is the same in every frame.
Due to the form of the potential energy in SM, in the minimum energy state we have a nonvanishing Higgs field. We have
$$
<\Omega|h(x)|\Omega>=v,
$$
where $h(x)$ is the Higgs field, $|\Omega>$ the vacuum state and $v$ an $U(2)$ matrix. This VEV is the same in every inertial frame.
I think that could be a little counterintuitive, so I'll go with an example. Let's go to electrostatics: here, you have a scalar potential $\phi$, that has the gauge freedom $\phi\to\phi+c$, where $c$ is a constant. Now, suppose that, for some wicked and totally unphysical reason, the energy of this potential is minimized for a certain value $\phi=d$, where $d$ is another constant. This choice of the potential surely breaks gauge invariance, but is still Lorentz invariant (well, static theory, Galileian invariant, but you can make a leap of immagination here). This is a simpler example of what you are asking: the nonvanishing VEV due to spontaneous symmetry breaking is a scalar in those cases, and you cannot use it to distinguish frames. Yet, it is not a scalar for gauge transformations, and you can say that you are choosing a particular gauge in order to minimize the energy.
