If you're sitting in the rocket then you appear to be stationary and it's the Earth that's moving. Therefore the people in the rocket will see time passing slowly on Earth. Of course we here on Earth see time passing slowly on the rocket.
The situation has to be symmetrical because it's a fundamental part of special relativity that all frames are equal and there are no special frames. If the people in the rocket saw time passing more quickly on Earth that would prove to them that they were moving and the Earth was stationary, and SR forbids this.
You need to be careful about tossing around concepts like time dilation as it's easy to fall into conceptual traps and end up with a paradox. The only reliable way to do things is to sit down with a piece of paper, choose the events you're interested in then do the Lorentz transforms to find out what happens.
Let me try and expand a bit on Ben's answer.
Starting with special relativity, the key thing to understand is that all the weird stuff, and indeed the Lorentz transformations, is derived from a property called the metric. If you have two points in spacetime separated by ($\mathrm dt,~\mathrm dx,~\mathrm dy,~\mathrm dz$) then the metric tells us how to calculate the interval between them. For SR this is:
$$ \mathrm ds^2 = -\mathrm dt^2 + \mathrm dx^2 +\mathrm dy^2 +\mathrm dz^2 $$
The interval $\mathrm ds$ is referred to as the line element and is an invariant, i.e., every observer no matter how fast they are moving, will calculate the same value for $\mathrm ds$.
The equation for the line element should remind you of Pythagoras' theorem, and indeed the only difference is that the sign of $\mathrm dt^2$ is negative not positive. It's this difference in the sign that is responsible for effects like time dilation. This is the important point to take home: this metric is all you need to calculate time dilation.
Now consider general relativity, and the effect of gravity. But first let me rewrite the special relativity equation for the line element in polar co-ordinates:
$$\mathrm ds^2 = -\mathrm dt^2 +\mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2\theta~\mathrm d\phi^2) $$
and now I'll write the equation for the line element near a black hole, i.e. the Schwarzschild metric:
$$ \mathrm ds^2 = -\left(1-\frac{2M}{r}\right)\mathrm dt^2 + \frac{\mathrm dr^2}{\left(1-\frac{2M}{r}\right)} + r^2 (\mathrm d\theta^2 + \sin^2\theta~\mathrm d\phi^2) $$
If you compare these two equations it should be immediately obvious that they are very similar, and indeed if you let the mass of the black hole, $M$, go to zero or if you go a long way away, so $r \rightarrow \infty$, then the two equations are the same.
This means the GR metric includes everything that the SR metric predicts, but it adds to it. So there isn't a distinction between the time dilation due to just velocity and the time dilation due to gravity. The GR metric is an extension of the SR metric and includes both. However let me reinforce Ben's cautions: it generally isn't useful to try and separate the time dilation due to velocity and the time dilation due to gravity.
Best Answer
Yes, both special relativity and general relativity have to be taken into account. The total time dilation is given by $$ \frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}}, $$ where $\text{d}\tau$ is the time measured by a moving clock at radius $r$, and $\text{d}t$ is the coordinate time measured by a hypothetical stationary clock infinitely far from the gravitational field.
For a person standing on the equator, we have $r_\text{eq}=6378\,\text{km}$ and $v_\text{eq}=0.465\,\text{km/s}$, and for a geostationary satellite $r_\text{s} = 42164\,\text{km}$ and $v_\text{s} = 3.074\,\text{km/s}$. This enables you to calculate $\text{d}\tau_\text{eq}/\text{d}t$ and $\text{d}\tau_\text{s}/\text{d}t$, and finally the ratio $\text{d}\tau_\text{eq}/\text{d}\tau_\text{s}$.
See also this post for more details: https://physics.stackexchange.com/a/90764/24142