What is the rigorous definition for current in a closed circuit? I know that $I = \frac {dq} {dt}$, but I have trouble reconciling that derivative definition with my intuitive understanding of current. At an arbitrary point in wire, the current is obviously the amount of charge flowing in and out per unit time -- but strictly speaking, wouldn't  $\frac {dq} {dt}$ model the rate of change of charge at that specific point, which should always be zero? 
It makes sense in the case of a capacitor, for example, where there's accumulation, but wouldn't the mathematical definition need to be slightly different to account for the balance of inflow/outflow by conservation of charge?
 A: You can think in terms of the current density vector. Its definition:
$$
\mathbf J = \sum_i n_i q_i \mathbf v_i
$$
Where $n_i$ is the charge carrier density in the media, $q_i$ is the charge of the charge carrier, and $\mathbf v_i$ is the average velocity of the charge carrier. Assuming we have several charge carriers (electrons, ions, etc), you have to sum they all to find the current density vector. 
Its clear now: The current density vector depends on the velocity of the charge carrier, the charge of it, and the density of it. Also note those are "volumetric" quantities as $n_i q_i = \rho_i$, which means the current density vector is a field: $\mathbf J = \mathbf J(\mathbf r, t)$. Then you can define the flux of it, in a given surface, and that's exactly a rigorous definition of the current $I$:
$$
I = \iint_S\mathbf J(\mathbf r, t)\cdot\mathbf{dS}
$$
In this sense, $I = dq/dt$ is the rate over time of charge being transported across this surface $S$. 

Since you also asked, I am now going to derive an equation for conservation of charge. With this definition of current, we have:
$$
I = \frac{dq}{dt} = \frac{d}{dt}\iiint_V \rho dV = 
\iint_V\frac{\partial\rho}{\partial t}dV
$$
The last step is only valid assuming a constant volume over time. If we think now in a closed surface $S$ where its interior is exactly the volume $V$, we can apply divergence theorem and came up with a conservation of charge:
$$
\iiint_V\frac{\partial\rho}{\partial t}dV = -\iint_S\mathbf J(\mathbf r, t)\cdot\mathbf{dS} = -\iiint_V\nabla\cdot\mathbf JdV
$$
Which basically means, the variation over time of charge density in the interior of $S$ is the current $I$ getting out across the boundary surface of $V$, that is, the surface $S$. We can now apply the divergence theorem:
$$
\iiint_V\left(\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf J\right)dV = 0
$$
And done! We have an equation for local conservation of charge as you asked, which takes account accumulation of charge and its transportation, which is called the continuity equation:
$$
\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf J = 0
$$
On global means, assuming $Q$ is the "acumulating" total charge inside $V$, and assuming $I$ to be the current getting out across surface $S$, that is, the boundary surface of $V$, the continuity equation becomes:
$$
\frac{dQ}{dt} + I = 0
$$
A: imagine a small section of the conductor, say 1mm thick...indeed on the left side, there is a given amount of charge exiting this section, but from the right side, there is an equal inflow of charge. For that reason, the total amount of charge in the section of conductor does not change
