Is the amount of cubic inches in a cubic foot exact up to infinite significant figures?
Dealing with imperial measures in science is problematic. The main reason for that is that you have (at least) two different definitions (fixed values) for a lot of them, one based on the laws of the United States and one based on the laws of the United Kingdom. While the differences are quite insignificant for everyday use (as in baking a cake), in the science world these small deviations can cause further inaccuracies. On Wikipedia you can find more, but for the argument, let's look at the values for the yard:
United States    0.914401829 m
United Kingdom   0.9143993   m
International    0.9144      m

In principle the definition of cubic inch to cubic foot is exact, while the conversion to the SI system may carry an error.
Unit        Abbreviation    Definition  SI equivalent (rounded 4sf)
Cubic inch  cu in or in³                16.38 cm³
Cubic foot  cu ft or ft³    1728 cu in  0.02831 m³
Cubic yard  cu yd or yd³    27 cu ft    0.7646 m³

Most of the time this shouldn't affect your measurements more than the use of significant figures, which is used as stating an error implicitly. That means it is in itself inaccurate. In any scientific capacity you would state the error explicitly.
When a value is stated as $1.00$ it has three significant figures it actually means that the measurement falls within the interval $0.99$ to $1.01$, or $1.00\pm0.01$.
Let's have a quick example. You measure the volume of water in a beaker to be $0.35~\pu{in^3}$. The beaker doesn't specify how accurate the measurement is, so we simply assume $\pm0.01~\pu{in^3}$. You would state that value as $0.35\pm0.01~\pu{in^3}$, or $0.34-0.36~\pu{in^3}$, or simply $0.35~\pu{in^3}$ with implied error. On the other hand, when we know significant figures are used, we would interpret a value of $0.35~\pu{in^3}$ as $0.35\pm0.01~\pu{in^3}$.
When you convert this to the SI system, your value doesn't become more or less accurate. You cannot use significant figures to express the error in one system in the same way you would do it in the other, because the implicit error is different.
The exact conversion (of the international yard) is
$$\frac{0.9144~\pu{m}}{1~\pu{yd}}=\frac{0.9144~\pu{m}}{3~\pu{ft}}=\frac{0.9144~\pu{m}}{36~\pu{in}}\implies1~\pu{in^3}=\left(\frac{0.9144}{36}\right)^3~\pu{m^3}.$$
The actual trust of the value is $\pm0.01~\pu{in^3}=\pm0.16387064~\pu{cm^3}$. When you convert your value $0.35~\pu{in^3}=5.74~\pu{cm^3}$ (keeping the same number of significant figures since the conversion is exact), you would imply an error of only $0.01~\pu{cm^3}$, which is far less than the actual trust. Instead your measured value falls somewhere between $5.58-5.90~\pu{cm^3}$. So the correct way of implicitly stating the error would be $6~\pu{cm^3}$.
TL;DR
The conversions are exact, the system of significant figures is not. In doubt, state error margins explicitly.