Relation between Gibbs energy and net reaction rate
Thank you for putting together an excellent, well-specified, thoroughly documented question.  The answer, as you suspected, largely comes down to what "small" and "large" mean.


  I have come across some papers in biochemistry that seem to claim that a reaction can be "near equilibrium" even though the flux through the reaction (net reaction rate) is large[....] How can this be? 


Consider the equilibrium approximation of Michaelis-Menten kinetics:

$$  \ce{E + S &lt;=&gt;[k_f][k_r] E\cdot S -&gt;[k_3] E + P}$$

When $k_r \gg k_3$, the enzyme-substrate complex $\ce{E \cdot S}$ is assumed to be in equilibrium with $\ce{E + S}$, i.e. free enzyme and free substrate.  Making this equilibrium assumption eventually provides the rate law for overall flux through the reaction:

$$\frac{dP}{dt} = k_3 E_0\frac{S}{S+\frac{k_r}{k_f}}$$

Note that the flux is not always zero even though we have assumed that $\ce{E\cdot S}$ is in equilibrium with $\ce{E + S}$!  This is one hint -- hopefully a familiar example if you have studied biochemistry -- that assuming (quasi-) equilibrium doesn't mean zero fluxes.


  In my understanding, the net rate of a reaction depends on the Gibbs energy $\Delta G$ so that, near equilibrium where $\Delta G \approx 0$, there is no net flux through the reaction. 


This isn't quite right.  First, you are right that the net rate of a reaction depends on the Gibbs energy $\Delta G$.  But near equilibrium where $\Delta G \approx 0$, it would be better to say "there is little net flux through the reaction".  Then the question is little relative to what?  The answer is that "little" is relative to the exchange flux.  So mathematically, 

$$\nu_{net} = \nu_f - \nu_r = k_f(E)( S) - k_r(E\cdot S) $$
$$\nu_{exchange} = \nu_r = k_r (E \cdot S) $$

Reactions that are "approximately" at equilibrium have $\nu_{exchange} \gg \nu_{net}$.  Thus, "little" is relative to the exchange flux, and not necessarily the rates of other reactions in a metabolic network.  


  ... "Small changes in $Q$ are accordingly adequate to tip the thermodynamically favored flux direction."


Remember that $\Delta G = \Delta G^\circ + RT \ln Q$.  The question of whether a reaction is reversible is asking about the sign of $\Delta G$.  The authors' statement that small changes in $Q$ can change the sign of $\Delta G$ only make sense if $\Delta G^\circ \approx 0$.  From the context of their remark, I think they are aware of this.


  But if this is true, the reaction direction seems unpredictable and the system should be difficult to control?


I wouldn't say "unpredictable" but I think you're right to say "difficult to control".  Another way of saying it would be that those reactions in the metabolic network that are approximately at equilbrium are not likely to be control points.  This 2006 paper examines that topic in more detail.


  Is this all just a question of what "small" or "large" $\Delta G$ means?


Yes!


  Can a reaction really be considered "near equilibrium" if the net reaction rate is large?


It depends what you mean by "large".  


If you mean "large" relative to some other flux in the network, then yes, a reaction can be near equilibrium even though its flux is "large".
If you mean "large" relative to the exchange/equilibration flux, then no, a reaction near equilibrium has a net flux that is very small compared to the exchange flux.