Simulation result of bjt cascode amplifier
As stated in my comment above, it seems to me that the problem of your circuit is that, at the fundamental frequency of the test input signal \$V_5\$, it behaves as a differentiator. I suggest to follow the steps below in order to see if it is really this the problem of the circuit, and if so solve it


A first check you could do is to increase the fundamental frequency \$f_{V_5}\$ of your input signal. From the analysis of your output waveform, it appears that \$f_{V_5}\simeq 130\mathrm{kHz}\$: try to raise progressively your frequency up to \$f_{V_5}= 500\mathrm{kHz}\$ and see what happens. If the output waveform resembles more and more a square wave, we have identified the problem and you can go to the second step
It seem to me that the biggest problem of your circuit is the low value of \$C_E\$ this implies an associated lower cutoff frequency of
$$
f_{c_\mathrm{Low}}=\frac{1}{2\pi R_E C_E}\simeq 59.0\mathrm{kHz}
$$
Try \$C_E=2700\mathrm{nF}\$ and see what happens. 


Further notes. 


While analyzing the influence of \$C_L\$, Harry Svensson simulated the response in the time domain of the high pass RC circuit that this capacitor makes with the load resistor \$R_L\$, to a square wave at \$125.0\mathrm{kHz}\$. The output waveform, shown in the comments to this answer, is distorted but resembles the shape of the input square wave: this confirms my intuition, supported intuitively by the fact that
$$
f_{R_LC_L}=\frac{1}{2\pi R_L C_L}\simeq 15.9\mathrm{kHz},
$$
that the problem lies in the value of \$C_E\$.
Tony EE rocketscientist, in his comments to the question, showed the simulation in the time domain of the response to a square wave at \$1.00\mathrm{MHz}\$ of an optimized cascode circuit: the shape of the output waveform looks very close to the input square wave, confirming again that a proper response of the circuit relies on a accurate design of the decoupling time constants.