Heat transfer can occur by conduction, by convection, and by radiation.
If you consider conduction through the bulk of the cup, the rate of heat transfer is directly proportional to the temperature difference across the material of the cup. As your experiment holds all variables equal except the temperature difference, cup A will lose heat at a faster rate than cup B, according to this version of Newton's Law of Cooling:
$$∆Q/∆t = −kA(∆T /L)$$
where $(∆Q/∆t)$ is the rate of heat conduction ($\mathrm{kJ/s}$), $∆T$ is temperature difference across the material, $L$ is thickness of the layer ($\mathrm{m}$), $A$ is area of the material ($\mathrm{m}^2$), and $k$ is thermal conductivity of the material per unit thickness ($\mathrm{kJ/m/s/°C}$).
The heat loss curve for cup A should be steeper than the curve for cup B. As you can see, the greater temperature difference ($∆T$) results in a greater rate of heat transfer from the cup to the ambient space.
Convection also plays a part in heat transfer. Convection is the movement of fluid inside the cup that circulates heat throughout the cup. The rate of heat transfer through convection is also directly proportional, to the temperature difference between the cup's surface and the fluid within the cup.
Another consideration is evaporative cooling at the liquid/air interface. The hotter the liquid, the faster it will evaporate and the more heat energy will dissipate per unit time. This is because hotter water molecules jiggle with greater energy, and more readily break the inter-molecular bonds that create surface tension. However, the rate of evaporation will increase only until the ambient air is saturated. The rate of evaporative cooling generally is dependent on more variables than conduction or convection, but it is directly proportional to temperature difference.
There is a reasonable chance that your tea is cooling faster.
This is somewhat counter-intuitive, but it's because the only variable here isn't the temperature.
Evaporation is also very important in the cooling of the tea. The kettle probably has much less available area to release the steam, so the air above is saturated. This will lead to a much slower rate of evaporation, compared to when the boiling water is exposed to open dry air. This would mean if you poured the same amount of water into each tea, yours would lose more mass through evaporation which could be the deciding factor.
The rate of evaporation depends on the surface area and vapour saturation in the air. A kettle has a small opening to exchange it's saturated air with the surroundings, while the open cup can easily move away the saturated air through natural convective currents and new fresh air can come in to take it's place.
If you put some sort of loose covering on your cup that emulated a kettle while it cools for the first few seconds (be careful to make sure steam can still get out, just not as easily) you should see it remain quite a bit warmer.
Edit: Adjusted the answer to reflect brought up in Vectorjohn's answer. I was getting too invested into evaporation and lost track of why I brought it up in the first place. It changes the mass of the tea you have to cool if you pour both to the same level.
Best Answer
The equation you give:
$$ Q = m C_p \Delta T $$
just tells us the total amount of heat transferred, and does not tell us anything about the rate at which the heat transfer occurs. To calculate heat flow we have to solve the heat equation. If you do a physics degree this apparently innocent equation will cause you many hours of frustrated head scratching, so in practice we tend to use simple approximations. In everyday life heat flow tends to be well described by Newton's equation:
$$ \frac{dQ}{dt} \propto \Delta T $$
so, as you suggest, the greater the temperature difference the faster the heat flow.
The experiment you describe isn't really a good way of showing this, because if you mix hot and cold liquid in practice the rate of temperature change will be controlled by how fast you do the mixing. Newton's equation would be more useful if you place the two liquids in contact but don't allow them to mix e.g. have a metal (or some other high thermal conductivity) divider between them. In that case you're quite correct that the initial heat flow will be faster at 100°C than at 70°C. However the 100°C system will take longer to cool because the amoutn of heat, $Q \propto \Delta T$, that needs to be transferred is greater with a 100°C difference. The temperature difference as a function of time will look like:
$$ \Delta T(t) = \Delta T_0 e^{-\alpha t} $$
where $\Delta T_0$ is the initial temperature and $\alpha$ is a constant related to the thermal conductivity (large $\alpha$ means high thermal conductivity). If I use this equation to graph the cooling for 70°C and 100°C initial temperatures (choosing a random value of $\alpha$) I get:
So even though the 100°C difference initially cools faster the 70°C difference always reaches any specified temperature difference before the 100°C does.