Current mirror and differential amplifier problem
Without more information to go on I would probably use the following approximation:

Looking at the voltages given, all BJT's are in the forward active region. This means that the collector will source/sink a more or less constant current, equal to \$\frac{I_0}{2} = 1mA\$. These current sources will shield the output voltage and current from everything happening elsewhere, so it does not influence \$\frac{I_{C1}}{I_{C2}}\$ in any meaningful way - as long as all transistors are in the active region.

The other way around does happen though. If a differential input voltage is applied, then \$I_{C1}\neq I_{C2}\$, and a non-zero current would flow through the resistor as you have stated (\$I_{out} \approx I_{C1} - I_{C2}\$). The difference between these currents will be proportional to Q1's and Q2's transconductance gain \$g_m\$, or

$$I_{C1} - I_{C2} \approx g_m\cdot (v_{B1} - v_{B2})$$

The output current will still be \$\approx I_{C1} - I_{C2}\$, and this is not influenced much by \$V_S\$ and \$R_C\$ because the output transistors still act like current sources.

The output voltage does change drastically though. If both base voltages are kept constant, then \$I_{out} \approx 0\$, and so \$V_{out} \approx V_s = 5V\$. Applying a differential voltage would mean that

$$\begin{align}
V_{out} &amp;\approx g_m\cdot (v_{B1} - v_{B2})\cdot R_C + V_S \\
&amp;= g_m\cdot (v_{B1} - v_{B2})\cdot 1k\Omega + 5V
\end{align}$$

If you would not connect \$R_C\$, then all current would be forced through the output impedance of the output transistors. These output impedances are generally very high for BJT's in the active region, so they can usually be neglected for \$R_C \ll Z_{out}\$. If this is not the case, you can replace \$R_C\$ with \$R_C || Z_{out}\$ to get a better approximation for the small-signal gain.