According to the Standard Model, were there 4 Higgs bosons at the beginning of the universe, before the electroweak symmetry breaking? My question is in the context of the Standard Model, not about Beyond Standard Model.
At the beginning of the Universe, before electroweak symmetry breaking, we have a complex Higgs isospin doublet. So could we say that at that period, there were 4 Higgs bosons?
After electroweak symmetry breaking, 3 Higgs fields (called Goldstone bosons) are eaten by the gauge bosons, and get mass, while the last Higgs field gets mass? So could we say that after electroweak symmetry breaking, there is one Higgs only?
But then, does it mean that the number of Higgs boson has varied between the Big Bang and the time at which the electroweak symmetry breaking occured?
 A: Whenever we say Higgs boson we actually mean "the quantum of the Higgs field, post SSB$^\dagger$". In the same way the "photon" is usually associated with the quantum of the post-SSB $A^\mu$ field, and not the pre-SSB (mixed) $B^\mu$.
The particles ( the quanta of the fields) are just excitations of the fields. The Higgs field operator $\hat{\phi}_H(\mathbf{x})$ will create one Higgs boson at position $\mathbf{x}$. The field itself, however, is the more imporant entity, and it permeates the whole of space.
If the field has $>1$ degrees of freedom, its quanta have $>1$ internal configurations.  The photon field $A^\mu$ has $2$ degrees of freedom: this corresponds to two mutually orthogonally polarised photons.
The $4$ degrees of freedom of the pre-SSB Higgs field can be thought as "polarisations" of the pre-SSB Higgs boson. Though this has imaginary mass so a physical interpretation is somewhat challenging (but let me know if you come up with one). 
I think they key problem with you is the Higgs field before and after SSB.


*

*Before SSB:
The Higgs field is a $SU(2)$ complex doublet, as you said, meaning it is a $2$-component quantity, each entry being complex:
$$ \phi_H = \left ( \begin{array}{c} \phi^+ \\ \phi^0 \end{array} \right ) =   \left ( \begin{array}{c} \phi^+_1 + i\phi^+_2  \\ \phi^0_1 + i \phi^0_2 \end{array} \right ),$$
with $\phi^+, \phi^0 \in \mathbb{C}$ and $\phi^+_1, \phi^+_2, \phi^0_1, \phi^0_2 \in \mathbb{R}$.  So there $4$ independent parameters, or "degrees of freedom" (d.o.f.). Each can be considered as a scalar field, yes -- scalar fields carry one degree of freedom each, so the maths checks out.
The Lagrangian density $\mathcal{L}$ has an $SU(2)\times U(1)$ symmetry, which has $4$ generators (different $4$ from before).  Upon requiring this symmetry to be local, you introduce $4$ gauge fields $W^1, W^2, W^3, B$. 
These have to be massless since bare mass terms for gauge bosons are not allowed in the Standard model, as they would violate gauge symmetry. A massless gauge field has $2$ degrees of freedom (two polarisations, think of a photon).
So: $4$ massless gauge fields ($8$ d.o.f.) $+$ $1$ complex Higgs doublet ($4$ d.o.f.) = $12$ d.o.f. 
Symmetry: $SU(2)_L \times U(1)_Y$.

*After SSB:
The Higgs mechanism happens. 
It breaks $SU(2)_L \times U(1)_Y$ into $U(1)_{\mathrm{em}}$. The number of broken generators are $4-1$ = $3$.
These would-be Goldstone bosons are "eaten up" by $3$ of the massless gauge fields, which now become massive. The Higgs mechanism also mixes these fields around, so now they are labelled $W^+, W^-, Z^0$ and $A$, the latter being the only one retaining its massless status (the photon field).
Massive gauge fields have $3$ degrees of freedom, think of an atom with spin $1$: it has projections $-1, 0, 1$.
In the process, the Higgs complex doublet reduces to a single real scalar field, what is usually referred as "the Higgs boson".
So: $3$ massive gauge fields ($9$ d.o.f.) + $1$ massless gauge field ($2$ d.o.f.) + $1$ real scalar field ($1$ d.o.f.) = $12$ d.o.f.
The number of degree of freedom of the $SU(2)_L \times U(1)_Y$ theory (electro-weak) remains unchanged. All that happens is a "reshuffling" around.
The Higgs boson pre-SSB, i.e. the quantum of the pre-SSB Higgs field, is a single particle carrying $4$ degrees of freedom ("polarisations"). Though it has imaginary mass so physical interpretation is challenging.
The Higgs boson post-SSB, i.e. the quantum of the post-SSB Higgs field, is a single particle carrying $1$ degree of freedom. This has a real mass term, and is what we associate with the experimentally observed particle.

$\dagger$ = Spontaneous Symmetry Breaking.
A: There is a confusion here, there is one Higgs boson in the standard model, which appears after symmetry breaking. Before symmetry breaking there is the Higgs field :

In the Standard Model, the Higgs particle is a boson with spin zero, no electric charge and no colour charge. It is also very unstable, decaying into other particles almost immediately. The Higgs field is a scalar field, with two neutral and two electrically charged components that form a complex doublet of the weak isospin SU(2) symmetry.
  bold mine.

So the particles with masses come after symmetry breaking. 
There is a complete distinction between particles and fields. There is everywhere a Higgs field on which creation and annihilation operators produce and annihilate the Higgs particle. Similar as for all particles in the table, for example: there is everywhere an electron field on which creation and annihilation operators produce and annihilate the electron. There is one electron.
Though there is a difference on the effect of symmetry breaking  in the model between fermions and bosons. In the standard model table, the fermions  exist before symmetry breaking with mass zero. The vector bosons are all within the multiplet of the unbroken gauge  "particle". At symmetry breaking energy in the cosmic models,  the fermions acquire mass through the higgs field, and the gauge bosons acquire mass and separate identities , as seen in the table, and separate fields on which creation and annihilation operators will work. The Higgs boson is a scalar particle , and it also gets its mass from the Higgs field afaik . See also this question.
