Like Borek said, the actual sign doesn't actually matter. In choosing a direction to go in, we're assuming that that is the direction of current! (That's why a resistor is always considered a voltage drop in going around, even though for our chosen direction the opposite may be true)
If we end up with a positive sign for the current once all is said and done, then our assumption was right. If however we get a negative sign, that means that we were wrong about the direction of current and that a minus sign is in order.
What matters is not the actual signs, but the relationships between the signs you assign to the different circuit elements in going around. As long as you use a consistent approach, imagining what a test particle in your pocket would experience as you go around, you'll get a correct result.
As for your comment about the battery, that's exactly what's so special about a battery!
We do need to invest energy to move across its terminals. But a battery is the source of the voltage and energy that's required! If you go across its terminals from (-) to (+), you are forcibly raised from low potential to high potential, you gain energy from the battery by going across!
As an exercise, let's now go in the other direction, again starting from the battery, and going CCW.
In going across the battery terminals we go across a voltage drop, but here's the tricky part: We are going against the direction of the field in the battery, so in fact, we have to invest energy to go across! Remember that a free charge would be accelerated by the battery from the (-) terminal to the (+) terminal, so in going the opposite way, we must overcome the force with which the battery tries to do this.
So our sign for the battery is -V_b
We now approach the negative capacitor plate and try to move up to the positive capacitor plate. To do that, we have to "climb up" a voltage, which means, to invest energy in doing so. So our sign for the capacitor is -V_c
And since we're overcoming the resistance of the resistor as we go across it, the sign is, as always, -IR
First let's go back to the CW result. For the CW result we found: V_b-IR+V_c=0
Rearranging:
CW: V_b+V_c=IR
And now we find: -V_b-V_c-IR=0
Rearranging:
CCW : V_b+V_c=-IR
Now you would come to me, and say, "But that's inconsistent!" since we got different signs for the current in each case.
But that would be overlooking the origin of each of those equations. A negative current CCW, is the same as a positive current CW!
Notice that we are dealing with positive quantities (V_b, V_c) and so we find that for our CW equation, IR is also a positive quantity. So the current I is also positive (This vindicates our choice of direction for the current).
For our CCW expression, however, we find that the sum of two positive quantities gives us a negative! How can that be? The only way out of the conundrum is to understand that we were wrong about the direction of the current and that in fact, it proceeds opposite the direction we marched in.
And so, we've shown that the two solutions are perfectly consistent as long as we're peachy with being wrong about the direction of current once.
A good way to avoid confusion over this issue is to assume a direction for the current and to draw it as a CCW/CW arrow inside the circuit loop and then use that as your meter stick in going around. So when you come to a resistor, you aren't always marching in the direction of current, you could be marching against your chosen direction, and actually experience a voltage increase in going across its contacts (+IR)!
Though I find it best to stick to one method so as not to become too confused. Find whatever works for you, either assuming an explicit direction for current, or going through the energy considerations for every component. With each of these you should be able to solve circuits quickly after some practice.
