Let's put together a simple mathematical model of an accelerometer - from this, we can work out some calibration options.
Ignoring non-linearity and other nasty effects, the output measurement of an accelerometer is given by:
$$\hat{\mathbf{f}} = \mathbf{M} \mathbf{f} + \mathbf{b}_a + \mathbf{n}_a$$
where \$\hat{\mathbf{f}}\$ is the actual measurement, \$\mathbf{b}_a\$ is the accelerometer bias, \$\mathbf{n}_a\$ is a random noise vector, \$\mathbf{f}\$ is the true specific force (i.e. acceleration) and \$\mathbf{M}\$ is the Scale Factor/Misalignment Matrix.
The individual elements of the SFA matrix are:
$$ \mathbf{M} = \begin{bmatrix} S_x && \gamma_{xy} && \gamma_{xz} \\ \gamma_{yx} && S_{yy} && \gamma_{yz} \\ S_x && \gamma_{zy} && S_{zz} \\ \end{bmatrix} $$
So, each scale factor is represented by an \$S\$ and each cross-axis sensitivity is represented by a \$\gamma\$.
Ideally, if the scale factor is 1 and there is no cross-axis sensitivity, then then the resulting matrix is \$\mathbf{M} = \mathbf{I}\$.
Representing it like this allows us to develop a compensation model. If we happen to know \$\mathbf{M}\$ and \$\mathbf{b}_a\$ and assume \$\mathbf{n}_a\$ to be small (i.e. close to zero), we can make a good estimate of the "true" acceleration from the measurements:
$$
\mathbf{f} = \mathbf{M}^{-1}\left(\hat{\mathbf{f}} - \mathbf{b}_a\right)
$$
The trick is, of course, working out \$\mathbf{M}\$ and \$\mathbf{b}_a\$.
I'll describe a procedure called the six position test, which is an easy and cheap way to calibrate an accelerometer. Step 1 is to mount the accelerometer in a rectangular box with perfectly \$90^\circ\$ sides (or as close as you can get). Place this on a perfectly level surface (or, again, as close as you can get) - you'd be surprised how good you can do this.
At this point, we know what the value should be: gravity on the z-accelerometer:
$$
\mathbf{f}_1 = \begin{bmatrix} 0 \\ 0 \\ g\end{bmatrix}
$$
So, this becomes:
$$
\hat{\mathbf{f}}_1 = \mathbf{M} \mathbf{f}_1 + \mathbf{b}_a + \mathbf{n}_a
$$
Noting that \$\hat{\mathbf{f}}_1\$ will close, but not the same as \$\mathbf{f}_1\$
If we put the box on it's head, the force acting is \$-g\$:
$$
\mathbf{f}_2 = \begin{bmatrix} 0 \\ 0 \\ -g\end{bmatrix}
$$
And when placed on one side:
$$
\mathbf{f}_3 = \begin{bmatrix} 0 \\ -g \\ 0\end{bmatrix}
$$
And so on for the remaining three sides.
Now, let's write out one of the equations longhand:
$$
\hat{\mathbf{f}}_1 = \mathbf{M} \mathbf{f}_1 + \mathbf{b}_a + \mathbf{n}_a
= \begin{bmatrix} S_{xx} f_x + \gamma_{xy} f_y + \gamma_{xz} f_z + b_x \\
\gamma_{yx} f_x + S_{yy} f_y + \gamma_{yz} f_z + b_y \\
\gamma_{xz} f_x + \gamma_{yz} f_y + S_{zz} f_z + b_z \end{bmatrix}
$$
And even longer hand (for the first one):
$$
\hat{\mathbf{f}}_1 = \begin{bmatrix}
f_x S_{xx} + f_y \gamma_{xy} + f_z \gamma_{xz} +
0 \gamma_{yx} + 0 S_{yy} + 0 \gamma_{yz} +
0 \gamma_{xz} + 0 \gamma_{yz} + 0 S_{zz} +
1 b_x + 0 b_y + 0 b_z \\
0 S_{xx} + 0 \gamma_{xy} + 0 \gamma_{xz} +
f_x \gamma_{yx} + f_y S_{yy} + f_z \gamma_{yz} +
0 \gamma_{xz} + 0 \gamma_{yz} + 0 S_{zz} +
0 b_x + 1 b_y + 0 b_z \\
0 S_{xx} + 0 \gamma_{xy} + 0 \gamma_{xz} +
0 \gamma_{yx} + 0 S_{yy} + 0 \gamma_{yz} +
f_x \gamma_{xz} + f_y \gamma_{yz} + f_z S_{zz} +
0 b_x + 0 b_y + 1 b_z
\end{bmatrix}
$$
So we can create a stacked vector of the unknowns
$$
\mathbf{z} = \mathbf{A} \mathbf{\beta}
$$
Where
$$
\mathbf{z} =
\begin{bmatrix}
\hat{\mathbf{f}}_1 \\
\hat{\mathbf{f}}_2 \\
\vdots \\
\hat{\mathbf{f}}_6
\end{bmatrix}
$$
And
$$
\mathbf{\beta} = \begin{bmatrix}
S_{xx} \\
\gamma_{xy} \\
\gamma_{xz} \\
\gamma_{yx} \\
S_{yy} \\
\gamma_{yz} \\
\gamma_{xz} \\
\gamma_{yz} \\
S_{zz} \\
b_x \\
b_y \\
b_z \\
\end{bmatrix}
$$
The design matrix is (for one set of measurements):
$$
\hat{A}_1 = \begin{bmatrix}
f_x && f_y && f_z &&
0 && 0 && 0 &&
0 && 0 && 0 &&
1 && 0 && 0 \\
0 && 0 && 0 &&
f_x && f_y && f_z &&
0 && 0 && 0 &&
0 && 1 && 0 \\
0 && 0 && 0 &&
0 && 0 && 0 &&
f_x && f_y && f_z &&
0 && 0 && 1
\end{bmatrix}
$$
Now, once this is setup, one may solve for \$\mathbf{\beta}\$ (and hence sensitivity and bias) via least squares.
A similar procedure may be performed with a robotic arm if you can precisely control the angles - it simply replies knowing the precise gravity at that angle which, if you know the angle, is easy to calculate.
Best Answer
First of all, the comment regarding your intended application is very important. For this answer, I'm going to assume the you want the highest data rate possible (i.e. Wii-type controller that sends near-instantaneous data). I give you two options below. In both cases, the receiver is simple: receiver board (either another XBee or the sister board of the transmitter you are using) connected to a USB-serial converter connected to your computer.
Option 1: Really easy to build, not-so-low power
An XBee 1mw Chip Antenna can be configured to transmit at whatever datarate you need, has a max. range of 100m, and, best of all, has six on-board ADCs. You can configure the XBee to automatically sample these and transmit their values. X, Y and Z from the accelerometer can go directly into AI0, AI1 and AI2 pins on the XBee and they can both be fed with 3.3V.
Option 2: Not so easy to build, low power
The transmitter board you mentioned can be powered by 3.3V (according to the datasheet). Alternatively, you could use this 2.4GHz transciever. In both cases, you will need a microcontroller (easy) or an ADC with serial output (harder, will need a PCB, something along the lines of the MAX1245) to convert the analog signals from the accelerometer to digital signals for the transmitter. I recommend you use a microcontroller (Arduino Pro Mini 3.3V for prototyping, then just an ATMega328 when you're done with the design).
Option 3: Stereo FM transmitters modulating 3 channels of data
This is in response to "Can I do it without converting the signals?".
On the receiver side:
NOTE: Most FM tuners will have band-pass filters that will screw this whole idea up.