Area of an hyperbolic triangle made by two geodesic and an horocycle Let D be the disk Poincare' model for the hyperbolic plane.
Let A be a point on the boundary $\partial D$ of the disk, consider two geodesics emanating from A and ending in other two points B and C of the boundary, in addition consider an horocycle centered in A and intersecting the two geodesics in P and Q.
Consider the region bounded by A, P and Q. What's its area?
If I were to use the usual formula for the area of a geodesic triangle I would get O becase the two geodesics meet with angle 0 in A, and meet the horocycle with angle $\pi/2$ at P and Q.
But is it possible for non empty region of D to have zero area?
 A: 
Consider the region bounded by $A$, $P$ and $Q$.

I assume that you mean the area bounded by two geodesics and one horocycle, or rather the corresponding portions thereof. Knowing the corner points doesn't strictly speaking tell you the path inbetween, but the problem statement strongly implies this, I'd say, and the question title seems to confirm this.

But is it possible for non empty region of D to have zero area?

No. You must not apply the regular angle deficit formula if your shape is not a polygon with geodesic edges. The horocycle makes the usual formula inapplicable.

What's its area?

In this specific example I would not use the Poincaré disk model, but instead the half plane model. Use the point at infinity as $A$, without loss of generality. Then the geodesics emanating from that will be vertical lines, and the horocycles centered around this will be horizontal lines. So you have to compute the area of a shape bounded by three Euclidean straight lines, namely two verticals and one horizontal. You can do that by integrating over the area element of this model, or by dividing this area into a number of triangles with true geodesics at the bottom, and then taking the limit as you increase the number of triangles.
Either way you will find that the result depends on your exact choice of boundary. Depending on which geodesics and which horocycle you choose, you will get a smaller or a larger area. There are instances of your setup which cannot be obtained from one another using a hyperbolic rigid motion, as can be seen e.g. from the number of real degrees of freedom this configuration has. Namely you have two real degrees of freedom for $P$ and $Q$ each, which will determine all the rest (up to some boolean decisions as you have two alternate choices for $A$). As a hyperbolic rigid motion has three real degrees of freedom, there have to be non-congruent configurations in there.
Does the problem statement provide any additional information?
For $P=(x_1,y_1)$ and $Q=(x_2,y_1)$ with $x_1<x_2$ in the half plane model the area can be computed as
$$\int_{x_1}^{x_2}\int_{y_1}^\infty\frac1{y^2}\,\mathrm dy\,\mathrm dx
=\int_{x_1}^{x_2}\frac1{y_1}\,\mathrm dx
=\frac{x_2-x_1}{y_1}\;.$$
If you compare this with the distance from $P$ to $Q$ along a geodesic (i.e. not following the horocycle) which is
$$\lvert P,Q\rvert=2\operatorname{Arsinh}\frac{x_2-x_1}{2y_1}$$
you can spot a striking similarity and use that to say the area in question is
$$2\sinh\left(\tfrac12\lvert P,Q\rvert\right)$$
which is a formula that can be applied independent of the model and the specific location of $A$, since hyperbolic distance is an intrinsic property of the hyperbolic plane which is independent from model and choice of coordinates.
A: The "usual formula" does not apply when the sides of a triangle are not geodesic. 
However, the "usual formula" is really just a special case of a more general formula known as the Gauss-Bonnet theorem, which does apply.
For your situation, the Gauss-Bonnet theorem says the following. Let $V$ vary over $A,P,Q$ and let $\angle_V$ denote the interior angle of $V$, which in the case of the infinite vertex $A$ is $\angle_A=0$. Let $\gamma$ vary over the three sides $AP$, $PQ$, $QA$ and let $\kappa_\gamma$ be the signed geodesic curvature. For a geodesic such as $AP$ or $QA$ we have $\kappa_\gamma=0$. The unsigned geodesic curvature of a horocycle equals $1$; when the outward normal is on the "outside" of the horocycle then the signed geodesic curvature is $\kappa_\gamma=+1$, whereas if the outward normal is on the "inside" then $\kappa_\gamma = -1$. In the case of the horocycle $PQ$, since it is centered at $A$ then the outward normal is on the "outside".
With all of that notation laid out, the Gauss-Bonnet theorem says
$$\sum_{V} (\pi - \angle_V) + \sum_\gamma \kappa_\gamma \cdot \text{Length}(\gamma) = 2\pi + \text{Area}(APQ)
$$
If all three sides were geodesics then this formula would simplify to the "usual formula"
$$\text{Area}(APQ) = \pi - \sum_V \angle_V
$$
However, in your case there is one more term,
$$\text{Area}(APQ) = \pi - \sum_V \angle_V + \text{Length}(\gamma) = \pi - \angle_B - \angle_C + \text{Length}(\gamma)
$$
Just as an example, in the upper half plane if we take $A=\infty$, $P=0+i$, $Q=0+i$ then the vertical lines $AP$, $AQ$ are geodesics, the horizontal line $PQ$ is a horocyclic segment centered on $A$, and this formula gives
$$\text{Area}(APQ) = \pi - \pi/2 - \pi/2 + 1 = 1
$$
which is consistent with what one obtains by integrating the upper half plane area form
$$\int_{x=0}^1 \int_{y=1}^\infty \frac{1}{y^2} \, dx \, dy
$$ 
