Can the current through this $6 \text{ }\Omega$ resistor be zero? In other words, can the $6 \text{ }\Omega$ resistor below have $I = 0$? I think the answer is yes, depending on the direction of the current.

Also, what do the negative $\epsilon$ values in the circuit signify? I think they signify which way the current is flowing?
 A: This comment and this other answer propose two long routes for the solution. Let me propose you a shorter one.
Denote by A and B the two nodes of your circuit. When $I=0$, it is indeed $V_\mathrm{AB} = \mathcal{E}$. Furthermore, when $I=0$, the branch with $\mathcal{E}$ and the $6\,\Omega$ resistor can be removed without changing $V_\mathrm{AB}$, and the voltage divider's formula yields
$$\mathcal{E}=V_\mathrm{AB} = \frac{3\,\Omega}{2\,\Omega+3\,\Omega}\times 28\,\mathrm{V} = 16.8\,\mathrm{V}.$$
A: Yes, an easy way to see this is to use  source transformation. Starting with your original circuit:

First, I'll transform your 28V source and series 2$\Omega$ resistor into a current source with parallel resistor:

Next, I'll combine the 2$\Omega$ and 3$\Omega$ resistors which are now in parallel (results in 6/5 $\Omega$ resistor):

Finally, transform back into a voltage source and you will see exactly what value of E your middle branch battery must be to result in no current in the branch of interest:

Using simple circuit analysis tools (source transformation, superposition etc.) can often greatly simplify the problem at hand.
I like @MassimoOrtolano's answer as the simplest presented for your particular case.
