Temperature and reaction quotient
Just to clear things up first (I might be saying something you already know, but I just want to be sure): the intent of the question is not to ask you what happens to the reaction quotient after equilibrium is re-established.

If a system is originally in equilibrium, with $Q = K_1$, and then you disturb it such that the equilibrium constant changes to $K_2$, then after the system re-reaches equilibrium you will have $Q = K_2$. If you interpret the question as asking whether $Q$ has changed after equilibrium is re-established, then the question basically amounts to asking whether the equilibrium constants $K_1$ and $K_2$ are different.

What the question is asking is whether $Q$ is affected instantaneously by the disturbance. Re-reading your question I think you got it, so...



Anyway, heat is not a product.

The reaction quotient for that reaction is defined as

$$Q = \frac{a_{\ce{CO2}}}{(a_{\ce{CO}})^2} \approx \frac{p_{\ce{CO2}}}{(p_{\ce{CO}})^2}$$

and $K$ is defined as the value of $Q$ at equilibrium.

Nowhere does heat appear in the equation. The idea of using heat as a "product" is nothing more than a mnemonic to help you remember the direction in which the equilibrium shifts. As long as the pressures are fixed, the temperature does not affect the instantaneous value of the reaction quotient.

When you increase the temperature, what happens is that the equilibrium constant $K$ drops. Let's say that before changing the temperature $Q = K = 0.01$ (just an example). If you increase the temperature, $K$ might fall to $0.001$ but $Q$ at that instant will still be $0.01$. As the system re-equilibrates, $Q$ will drop to $0.001$.

So, it is not that $K$ remains constant and $Q$ increases. It is more like: $K$ drops and $Q$ remains constant (at least instantaneously).