Neither. Information isn't energy and can't be measured using energy units. As you pointed out, information is related to entropy which is about degrees of freedom in a system.
In physics, information is usually measured in nats but bits are common too.
This is more a mathematical problem than a physical one.
As I understand, you accept that $ \oint_\gamma \frac{\delta Q}{T} = 0$ if $\gamma$ is a Carnot cycle.
Now, any cycle in the thermodynamic state space can be approximated with fragments of Carnot cycles with an aritrary precision. It can be shown that for such cycles you will also have $ \oint_\gamma \frac{\delta Q}{T} = 0$ because the region surrounded by the graph can be divided into many subregions that are surrounded by Carnot cycles, and $\oint_\gamma \frac{\delta Q}{T} = \sum_i \oint_{\gamma_i} \frac{\delta Q}{T}$.
Since the precision of approximation can be as good as we want, by continuity we can prove that $ \oint_\gamma \frac{\delta Q}{T} = 0$ for any cycle.
Now, consider two curves (processes) going from point A of the state space to the point B, let's call them $\gamma_1$ and $\gamma_2$. If you follow one of the curves and then backtrack the other, you get a cycle $\gamma$. You have
$$ 0 = \oint_\gamma \frac{\delta Q}{T} =\int_{\gamma_1} \frac{\delta Q}{T} -\int_{\gamma_2} \frac{\delta Q}{T} $$
that is
$$ \int_{\gamma_1} \frac{\delta Q}{T} =\int_{\gamma_2} \frac{\delta Q}{T} $$
The curves $\gamma_1$ and $\gamma_2$ were chosen arbitrarily, which means that the integral $\int_\gamma \frac{\delta Q}{T}$ over a curve (process) $\gamma$ linking two states $A$ i $B$ does not depend on the specific of the process; it can only depend on the starting and ending points.
Let us choose an arbitrary state $O$. For $A$ being another state, let us define a function
$$ S(A) = \int_O^A \frac{\delta Q}{T}$$
where the integration is performed over any curve linking state O to state A (as w ehave proven, it doesn't matter which one). This is what we called entropy. By construction, it's a function of state.
It can also be proven that $$ \Delta S_{AB} = S(B)-S(A) = \int_A^B \frac{\delta Q}{T}$$
where the integration is over any process from state A to state B.
There is some arbitrariness in the definition of the entropy function (choosing state $O$) but choosing a different state $O'$ would only change the entropy function by a constant.
Best Answer
What you're looking for is Landauer's principle. You should be able to find plenty of information about it now that you know its name, but briefly, there is a thermodynamic limit that says you have to use $k_\mathrm BT \ln 2$ joules of energy (where $k_\mathrm B$ is Boltzmann's constant and $T$ is the ambient temperature) every time you erase one bit of computer memory. With a bit of trickery, all the other operations that a computer does can be performed without using any energy at all.
This set of tricks is called reversible computing. It turns out that you can make any computation reversible, thus avoiding the need to erase bits and therefore use energy, but you end up having to store all sorts of junk data in memory because you're not allowed to erase it. However, there are tricks for dealing with that as well. It's quite a well-developed area of mathematical theory, partly because the theory of quantum computing builds upon it.
The energy consumed by erasing a bit is given off as heat. When you erase a bit of memory you reduce the information entropy of your computer by one bit, and to do this you have to increase the thermodynamic entropy of its environment by one bit, which is equal to $k_\mathrm B \ln 2$ joules per kelvin. The easiest way to do this is to add heat to the environment, which gives the $k_\mathrm BT \ln 2$ figure above. (In principle there's nothing special about heat, and the entropy of the environment could also be increased by changing its volume or driving a chemical reaction, but people pretty much universally think of Landauer's limit in terms of heat and energy rather than those other things.)
Of course, all of this is in theory only. Any practical computer that we've constructed so far uses many orders of magnitude more energy than Landauer's limit.