negative current for discharging capacitor
I once had the same doubt, but in short, it has to do with the passive sign convention.

This is the circuit that you have:







simulate this circuit &ndash; Schematic created using CircuitLab

See that instead of using KVL, I am using KCL for now. I defined the node \$v_o\$. I have defined my currents in the direction shown, but you can certainly choose other directions. 
It follows that: 

$$ i_c + i_R = 0$$

And you could now plug in what \$i_c\$ and \$i_R\$ are, to get 

$$C\dfrac{dv_o(t)}{dt}+\dfrac{v_o}{R}=0$$ 

And that's the differential equation that will give you the well known solution for a discharging capacitor. 

Why does it work out for KCL and you can't seem to get it to work using KVL?

The trick is in the use of the positive sign convention. Passive devices have a positive current and voltage relationship when the 'current is going into the positive terminal and comes out of the negative terminal'

Since the current is going into the elements through the + terminal and comes out through the negative terminal then the current is positive, by the PSC. 

Here is an excerpt from Nilsson-Riedel Electric Circuits book 



So when you see the capacitor has the \$i_c=+C\dfrac{dv}{dt}\$ (notice the +), that's the definition following the passive sign convention where the current enters the positive terminal. 

If you were to use KVL, take a look a the following approach:







simulate this circuit

Which is the same as the first one I drew but I added another current defintion, the one I will use for the KVL loop. I named that current \$i_s\$ (in red), and let's do KVL:

$$ v_o-i_sR=0$$

That's where you get confused. Now, you can see from the circuit that \$i_s\$ goes in the opposite direction compared to the \$i_c\$ current (definition by psc), that is, 

\$i_c=-i_s\$ or \$-i_c=i_s\$

And since the current \$i_s\$ is in the same direction as \$i_r\$,

$$i_s=i_r $$

And if you plug the \$i_s\$ and \$i_c\$ relationship, you end up with the right differential equation:

$$ v_o-(-i_c)R=0$$
$$ v_o+i_cR=0$$
$$C\dfrac{dv_o(t)}{dt}+\dfrac{v_o}{R}=0$$ 

Hope it helps.