What will happen after escaping earth's gravitational field? Suppose that I escaped the gravitational field of earth.
Then: am I going to be pulled by Sun's gravity?
 A: Yes, you will be pulled by the sun's gravity.  However, that has nothing to do with whether you have "escaped" earth's gravitational field or not.  That is a non-sensical concept because the gravitational fields of objects don't have a hard distance limit where they suddenly go to zero.  Once you get far enough from a object, the gravitational pull from that object will fall off with the square of the distance.
Therefore, no matter where you are or how far you go, you will always be within the gravitational field of the earth, the sun, and any other piece of matter.  The only issues is that if you go far enough, the forces from those objects will be so small that you can usefully approximate them as zero.  However, that threshold varies by application, so again there is no hard distance where you can even pretend the earth's gravitational field is irrelevant.
A: When at distance $R$ from the center of earth, the minimum velocity needed to escape from earth's gravity is $v_{esc} = \sqrt{2GM/R}$. Here, $G$ denotes Newton's gravitational constant, and $M$ earth's mass. 
As should be obvious from this equation, regardless the distance $R$, the escape velocity never vanishes. So no matter how far you travel, you never escape earth's gravity! 
Some figures to put things in perspective:
Using $2GM = 7.9 *10^{5} m^3 s^{-2}$, it follows that at one earth radius distance ($6.37 *10^6 m$) from earth's center (i.e. at earth's surface), the escape velocity equals $11 km/s$. 
You have to fourfold this distance (be four earth radii away from earth's center) to half this escape velocity. Even when being a moon's distance (60 earth radii) away from earth, the escape velocity still is more than $1/8$ of that at earth's surface.

Although regardless your distance your escape velocity never falls to zero, you might define the 'escape distance' as the distance beyond which the Hubble expansion $H_0 R$ exceeds the escape velocity $\sqrt{2GM/R}$. Physically, this corresponds to the expansion of space trumping the gravitational attraction. It follows that the escape distance thus defined is given by $R_{esc} = {(2GM/H_0^2)}^{1/3}$. 
Using $H_0 = 1.62 * 10^{-18} s^{-1}$, we find for earth an escape distance equal to $6.7 * 10^{16} m$ or $7$ lightyears. A considerable distance given that no human has travelled farther from earth than 1.3 light seconds!
A: Mass of the Sun is $1.9891\times10^{30}\approx2\times10^{30}kg$ $_1$. Lets consider your mass to be $100$ kilograms, just for convenience.   
Gravitational constant is about $6.62\times10^{-11}Nm^2kg^{-2}$. Distance of the Sun from earth is about $149,600,000\approx1496\times10^{5}km$ $_2$. Then, your distance from the sun can be assumed to be about $149\times10^{6}km$.
Then, attraction between you and sun will be about$_3$ $$F=G\frac{Mm}{r^2}=6.62\times10^{-11}Nm^2kg^{-2}\frac{2\times10^{30}\times10^{2}kg^2}{(149\times10^6\times10^3m)^2}=0.6N$$    
So, you will be accelerated towards sun with the acceleration of $0.6/10^{2}(=0.006ms^{-2})$.    
Remember you can't escape from earths gravitational field, it extends to infinity.
Overall it depends on in which direction you are projected. The world is complex as commented by others and is not easy as we think, I made this calculation just to show you how  complex the universe is. You should also consider the force towards earth due to earth and other planets and all other factors, which is not considered here. 

Credits: $_1$Google$_2$Google$_3$GoogleCalci
