Definition of the stress-energy tensor Why don't we denote presence of matter and energy in space-time (as we do for presence of electric charge) by a 4-vector field?
Is there any information in the stress-energy tensor that is not in 4-vector field of current of energy and matter? As I think there is a current of energy and matter about us very similar to current of charges, and we denote this by 4-vector field $J$.
I hope there is explanation as this example: if someone ask me why we don't denote the presence of charge about us just by scalar field of charge density, then I say charge density of course denote the presence of charge but it neglects current of charges and there are physical phenomenons that arise not from only charges but also from their motion.
 A: 
Why don't we denote presence of matter and energy in space-time (as we do for presence of electric charge) by a 4-vector field?

Because energy is not Lorentz-invariant, whereas electric charge is. Therefore, when we compute how charge density and its flux transform to assert that charge density together with its flux is a four vector, we only need to account for the fact that different observers see the partitions of spacetime used to calculate the density and its flux differently, they do not see the charges themselves differently. The Lorentz covariance of the 4-current is equivalent to an assertion that electric charge is a Lorentz-invariant scalar. 
In contrast, energy is not a Lorentz covariant scalar. It is united with momentum in the 4-momentum. So when we calculate the density of this entity, we need to account for the fact that both the underlying entity and the spacetime partitioning used to calculate a density change for different observers. We are computing the density and flux of a rank-1 tensor, not of a scalar.
This is one reason why Gravitoelectromagnetism doesn't work, although it gives pretty accurate calculations in many relativistic cases. It assumes, like Maxwell's equations, that the gravity field source is a rank 1 tensor - the mass/energy density and its flux - but this is flawed because mass / energy is not Lorentz invariant. Gravitoelectromagnetism gives the wrong values of gravitational radiation - the Larmor formula and its relativistic equivalent foretell radiation quantities that are far too big (falsified by observations of the Hulse-Taylor binary star, for example).

Is there any information in the stress-energy tensor that is not in 4-vector field of current of energy and matter? As I think there is a current of energy and matter about us very similar to current of charges, and we denote this by 4-vector field J
  J.

I think I've answered this one. Essentially the rank 2 stress energy tensor accounts for the Lorentz transformation of both the underlying quantities whose density is being calculated as well as the Lorentz transformation of the spacetime partitions used to calculate the density: a rank 1 four vector can only encode the Lorentz transformation of the latter.
A: "Canonical" reasoning:
In a field theory, associated to every one-parameter family of symmetries is a current 4-vector, which is conserved in terms of being divergenceless.
Now, since Minkowski spacetime is 4 dimensional, there are 4 linearly independent translations: 3 spatial and one temporal.
From nonrelativistic mechanics, we know that momentum is the Noether-charge associated to spatial translations. Since there are 3 spatial translations, momentum has three components. But these are Noether charges, not currents. This means there are three Noether currents associated with spatial translations, $\mathcal{J}_1^\mu,\mathcal{J}_2^\mu,\mathcal{J}_3^\mu$, where $\mathcal{J}^0_i$ denotes the momentum density in the ith direction, and $\mathcal{J}^j_i$ denotes the $j$th component of the momentum flux in the ith direction. The $\sigma_{ij}=\mathcal{J}^j_i$ components form a 3-tensor that is the ordinary stress tensor (or at least related to it by a constant).
If we also consider temporal translations, we also get $\mathcal{J}^\mu_0$, which is the Noether current of time translations. The $\mathcal{J}^0_0$ component is the energy density, and the $\mathcal{J}^i_0$ component is the ith component of the energy flux. The components $\mathcal{J}^\mu_\nu$ together form the $\Theta^\mu_{\ \nu}$ canonical stress-energy tensor. This makes sense only in flat Minkowski spacetime, and these components do transform tensorially.
This tensor can be symmetrized (the Belinfante-Rosenfeld tensor is one such symmetrization), and it can be proven (look around the site for related questions and answers/links therein) that the Belinfante tensor (which is equivalent modulo gauge transformations to the canonical tensor, in particular the conserved charges agree) is equivalent to the general-relativistic Hilbert stress-energy tensor in the flat spacetime limit.
So 1 symmetry $\Longrightarrow$ 4 current components, 4 symmetries $\Longrightarrow$ $4\times 4=16$ current components, hence it is why a second order tensor.
General relativistic reasoning:
The metric tensor naturally couples to basically any sort of matter fields, because the invariance of the action necessitates contractions between derivatives of fields ($\partial_\mu\phi\partial^\mu\phi$ type of stuff), plus the volume element $\sqrt{-g}$ depends on the metric too. The exceptions are topological fields, which have no local degrees of freedom.
Which means that if one forms the total action for a gravitational system, $$ S_t=S_{EH}+S_m,$$ where $S_{EH}=\int R\sqrt{-g}d^4x$ and $S_m$ is the action for matter fields, $S_m$ will be a functional of the metric too, not just the matter fields. To get the (gravitational) field equations, we need to functionally differentiate with respect to $g^{\mu\nu}$, which gives $$ \frac{\delta S_{EH}}{\delta g^{\mu\nu}}=-\frac{\delta S_{m}}{\delta g^{\mu\nu}}. $$ Because $g$ is symmetric, the RHS must be a symmetric second order tensor, which we identify as $\frac{1}{2}T_{\mu\nu}$, so the stress-energy tensor is a second-order tensor because the field variable (the metric) is also second order. Second order configuration variable $\Longrightarrow$ second order source current.

Edit: To elaborate the "canonical reasoning":
You used an example of charges: To charge $Q$, we associate a charge density function $\rho$, and a charge current $\mathbf{j}$. The continuity equation is $$\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{j}=0.$$
Relativistically, the charge density and the current form a 4-vector $j^\mu=(\rho,j_x,j_y,j_z)$ (I set natural units). The continuity equation is $\partial_\mu j^\mu=0$.
What about energy and momentum? To energy $E$ we associate a density $u$ and a current vector field $\mathbf{S}$. The continuity equation is $$ \frac{\partial u}{\partial t}+\nabla\cdot\mathbf{S}=0.$$
But momentum is itself a 3-vector, but this 3-vector plays the role of a charge, not a current. Consider component-wise, the $x$-component of momentum is $p_x$. We associate to it a momentum density $\Pi_x$, and a momentum current $\vec{\sigma_x}$. The continuity equation is $$ \frac{\partial\Pi_x}{\partial t}+\nabla\cdot\vec{\sigma_x}=0. $$
If we consider all 3 momentum components we get a "higher order" continuity equation $$ \frac{\partial\vec{\Pi}}{\partial t}+\nabla\cdot\sigma=0, $$ where $\sigma$ is now a second order tensor.
But we know from relativity, that energy and momentum (the charges!) together form a 4-vector $p^\mu=(E,p_x,p_y,p_z)$, which implies that the current is a $4\times 4$ component object $$ \Theta^\mu_{\ \nu}=\left(\begin{matrix} u & S_x & S_y & S_z \\ \Pi_x & \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \Pi_y & \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \Pi_z & \sigma_{zx} & \sigma_{zy} & \sigma_{zz}  \end{matrix}\right). $$
A: There is an explanation, and it's relatively simple. But it does have deep consequences.
We can in fact, and do define the stress energy tensor in general relativity (GR) by $T^{\mu\nu}$, and it represents all contributions to the sources of gravity except for the energy-momentum (and stress) of the gravitation itself. The tensor is a covariant entity, and it includes terms for the energy, momentum, and stress of mass-energy. But it does not include the contribution from the gravitational effect. This has significant consequences - for one, that energy-momentum is NOT generally conserved in GR. 
In the Einstein GR equations 
$G^{\mu\nu} = kT^{\mu\nu}$,
the G tensor is the gravitation, actually also some terms having to do with the curvature of spacetime, and the equation says that matter-energy (on the right) causes curvature of spacetime. But the G tensor it turns out is not a linear function of the metric, and the contribution of nonlinear (and non trivially characterized) components of G can carry energy and momentum.
And in general then, the energy-momentum from the T tensor is not conserved. The ordinary divergence of the T tensor is not zero, so there is no conserved current. That is 
Ordinary Div (T) = $\partial_{\nu}T^{\mu\nu}$ is not equal to zero
What turns out to be true is that the covariant divergence is indeed zero, 
Covariant Div (T) = 0
The covariant divergence, formed from the covariant derivative, is zero. But you can't turn it into a partial derivative or ordinary divergence so you can't integrate on the boundary to get zero using Gauss's law. Covariant derivatives include terms formed from derivatives of the metric, that is the connection or Christoffel symbols. That includes the nonlinear effect of gravity in GR. 
And energy and momentum is not generally conserved in GR. It is in some special spacetimes, such as where there is a global time symmetry, and where the spacetime is asymptotically flat (and then only total energy, not localized energy). 
See for instance the energy stress tensor in GR at https://en.m.wikipedia.org/wiki/Stress–energy_tensor
It does run out that one can manipulate the covariant divergence equation and define another entity, a pseudotensor $\tau$, such that the 
Ordinary Div ($T + \tau$) = 0
The pseudotensor represents the stress energy of gravitation, i.e. of spacetime. One could then try to say that the energy momentum vector defined from this is conserved. But It is unfortunately not a covariant entity, it is different in different coordinate systems, and there is no unique definition possible of it. But it can be used in some cases. See towards the end about mass is GR.
See the gravitational stress energy pseudotensor at https://en.m.wikipedia.org/wiki/Stress–energy–momentum_pseudotensor. As you can see form there, there is no unique expression for it, there are various that have been used, and they are not covariant Tensors. Each gives different results in different coordinate systems. Only when you have special spacetimes as indicated above does energy momentum conservation holds or makes sense. In cosmological spacetimes it does not. 
So, the bottom line is that in GR there is a stress energy tensor that represents the energy momentum (and stress) of matter energy, except for gravitational mass-energy. And it is of course not conserved. And the gravitational mass energy is not a covariant entity, nor really unique. 
The same is true of mass in GR. It is the energy, and it is not unique in GR. You can see various useful definitions of mass for special spacetimes at https://en.m.wikipedia.org/wiki/Mass_in_general_relativity
That is the impact of gravitation being spacetime curvature. 
