I'll make an example, to make things clear.
Take a two body system, in which the particles are seperated by a constant distance $d$ and have mass $m_1 = m_2 = m$.
This is a holonomic constraint, since
$$ | \vec{r}_1 - \vec{r}_2 | = d $$
with the particle-positions $\vec{r}_1$ and $\vec{r}_2$.
This system is therefore reduced to 5 degrees of freedom (6 minus 1).
Let's assume that we wanna look at this system when it is initially at rest and the applied external forces will always point perpendicular to the axis of rotation. This means that the system's movement will be constrained to a plane. Let's take a coordinate system such that this movement is in the x-y-plane.
We have therefore 3 effective degrees of freedom (4 minus 1 constraint). One is allowed to choose arbitrary independent variables $q_j$ corresponding to these degrees of freedome.
Take $q_1 := x$ the x-coordinate of the center of mass, $q_2 := y$ the y-coordinate and $q_3 := \phi$ the angle of rotation of the vector joining the two particles relative to the x-axis
$$ \cos\phi = \frac{\vec{d}\cdot\vec{e}_x}{d} $$
with $\vec{d} = \vec{r}_1 - \vec{r}_2$ and $\vec{e}_x$ the unit basis vector pointing along the x-axis. Then we have:
\begin{align}
\vec{r}_1 & = \underbrace{\left( \matrix{x\\y} \right)}_{:=\vec{r}_c} + \underbrace{\frac{d}{2} \cdot \left( \matrix{\cos\phi \\ \sin\phi} \right)}_{=\vec{d}/2} \\ \\
\vec{r}_2 & = \left( \matrix{x\\y} \right) - \frac{d}{2} \cdot \left( \matrix{\cos\phi \\ \sin\phi} \right)
\end{align}
The equations of motion are given by d'Alembert's principle, which states that the internal forces of constraints do no net work on the body:
$$ \sum_{i=1}^2 \left( m_i \frac{d^2\vec{r}_i}{dt^2} - \vec{F}_{i,\text{ext}} \right) \cdot \frac{\partial \vec{r}_i}{\partial q_j} = 0$$
which gives for every $j=1,2,3$ one equation of motion. For $q_1 = x$ we have:
\begin{align}
& \sum_{i=1}^2 \left( m_i \frac{d^2\vec{r}_i}{dt^2} - \vec{F}_{i,\text{ext}} \right) \cdot \left( \matrix{1\\0} \right) = \left( \sum_{i=1}^2 \left( m_i \frac{d^2\vec{r}_i}{dt^2} - \vec{F}_{i,\text{ext}} \right) \right) \cdot \left( \matrix{1\\0} \right) \\
& = \left( m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{1,\text{ext}} + m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{2,\text{ext}} + m \frac{d^2(\vec{d}/2)}{dt^2} - m \frac{d^2(\vec{d}/2)}{dt^2} \right) \cdot \left( \matrix{1\\0} \right) \\
& = \left( 2m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{1,\text{ext}} - \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{1\\0} \right) = 0
\end{align}
And similar for $q_2 = y$:
$$ \left( 2m \frac{d^2\vec{r}_c}{dt^2} - \vec{F}_{1,\text{ext}} - \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{0\\1} \right) = 0 $$
Or equivalently to accord for both equations:
$$ M \frac{d^2\vec{r}_c}{dt^2} = \sum_i \vec{F}_{i,\text{ext}} \tag1$$
the equation of motion of the center of mass $\vec{r}_c$ with $M = 2m$. This shows, that the motion of the center of mass is not affected by the point on which the force is applied (e.g $\vec{F}_{1,\text{ext}} = \vec{F}_0$ and $\vec{F}_{2,\text{ext}} = 0$ leads to the same motion as $\vec{F}_{1,\text{ext}} = 0$ and $\vec{F}_{2,\text{ext}} = \vec{F}_0$). The center of mass moves according to the external forces that are applied. This is a general result, which was for instance answered here:
Equation of motion for the center of mass of a rigid body
Now let's take the last equation for $q_3 = \phi$:
\begin{align}
& \left( m \frac{d^2\vec{r}_c}{dt^2} + m \frac{d^2(\vec{d}/2)}{dt^2} - \vec{F}_{1,\text{ext}} \right) \cdot \left( \matrix{-\sin\phi \\ \cos\phi} \right) \\
& + \left( m \frac{d^2\vec{r}_c}{dt^2} - m \frac{d^2(\vec{d}/2)}{dt^2} - \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{\sin\phi \\ -\cos\phi} \right) \\
& = \left( m \frac{d^2(\vec{d})}{dt^2} - \vec{F}_{1,\text{ext}} + \vec{F}_{2,\text{ext}} \right) \cdot \left( \matrix{-\sin\phi \\ \cos\phi} \right) = 0
\end{align}
this is independent of $\vec{r}_c$ and therefore a pure differential equation for $\phi$, which describes the rotational movement. We see that, the rotation around the center of mass is independent of its linear motion, in the sense that it can be described by equations that ignore the movement of the center of mass. One can express that in terms of torque and angular momentum $\vec{L}_r$ relative to the center of mass
$$ \frac{d \vec{L}_r}{dt} = \sum_i \vec{d}_i \times \vec{F}_{i,\text{ext}} \tag2 $$
with $\vec{L}_r = \Theta \cdot \vec{\omega}$ and $\Theta$ the inertia tensor and $\vec{\omega}$ the angular velocity (e.g. $\frac{d\phi}{dt}$ in the above example) at the center of mass and $d_i$ the distances to it.
Regarding your energy problem, I refer you to the last section of this answer:
https://physics.stackexchange.com/a/174208/75518
or think about it like this: Work is defined by the path-integral $\int \vec{F} \cdot d\vec{r}$ or for a many body, rigid system, since the net work of internal forces is zero:
\begin{align}
\sum_i \int \vec{F}_{i,\text{ext}} \cdot d\vec{r}_i & = \sum_i \int \vec{F}_{i,\text{ext}} \cdot \frac{d\vec{r}_i}{dt} dt \\
& = \int \sum_i \left( \vec{F}_{i,\text{ext}} \cdot \frac{d (\vec{r}_c + \vec{r}_{i,\text{rot}})}{dt} \right) dt \\
& = \int \left( \sum_i\vec{F}_{i,\text{ext}} \right) \cdot \vec{v}_c~dt + \int \left( \sum_i\vec{F}_{i,\text{ext}} \cdot \vec{v}_{i,\text{rot}} \right)~dt \tag3
\end{align}
where we have split the motion of a particle into the linear motion of the center of mass, which every particle shares, plus the rotational motion. In the above example this is $\vec{r}_{i,\text{rot}} = \pm \frac{\vec{d}}{2}$. As you can see, the first term corresponds to the work done on the center of mass, while the second corresponds to the work done to let the system rotate. It is this extra displacement $d \vec{r}_{i,\text{rot}}$ that comes with rotation, that causes the system to gain more kinetic-energy in the same ammount of time from a given force of equal magnitude.
Conclusions:
- "Will the body still accelerate as much as it did when F was applied on C?" $\rightarrow$ yes, the acceleration of the center of mass is the same (see equation (1))
- "If so, why?" $\rightarrow$ because the movement of the center of mass is independent of the point where forces are applied. Only the net force matters (see equation (1))
- "How can it rotate and yet accelerate with the same velocity as that of without rotation?" $\rightarrow$ because those two motions are independant of each other
- "Where is it getting this extra energy from?" $\rightarrow$ last section of https://physics.stackexchange.com/a/174208/75518 or see equation (3)
- "How can I calculate the respective velocities of linear and angular motion given F and t?" $\rightarrow$ solve equations (1) and (2) for specific forces.
Note: An impact on a rigid body is, however, a total different situation. See this: Elastic collision of point particle and rod
