The basic problem is you are getting hung up on irrelevant details.
The filter rolloff frequency is 10.3 Hz, and your input frequency is 300 Hz. That means the input frequency is 29x the rolloff, or 4.9 octaves. That's "a lot", so the output amplitude is going to be very close to the input amplitude. Put another way, the filter gain will be 1.0 for practical purposes.
The expected filter gain is 1.0, and you're getting 1.00. What's the problem? You're upset because a numerical simulation is off by a fraction of a percent? Really!? Are you really going to implement this filter with a .1% resistor and capacitor?
The transfer function is: $$\small G(s)=\frac{1}{(RC)^2s^2+3RCs+1}$$
Hence \$\omega_n=\frac{1}{RC}=\small10^4\:rad/s\:(=1592\:Hz)\$, and \$\small\zeta=1.5\$, and it can be seen that the DC gain (\$\small s=0\$) is unity.
Converting this to the frequency domain, using \$ s\rightarrow j\omega\$:
$$\small G(j\omega)=\frac{1}{1-(\omega RC)^2+j3\omega RC}$$
At \$\small 10\:\small Hz\$, \$\small \omega RC=0.00628\$, hence the gain is almost unity and the phase angle is almost zero. At \$\small 1\:\small kHz\$, \$\small \omega RC=0.628\$, giving a gain of \$\small 0.505\$, and phase angle of \$\small \phi=-72^o\$.
So it seems that there's a problem with your experimental set-up. What's the input impedance of the instrument measuring Vout?
Let's do some detective work:
If the input impedance of the instrument were \$\small 3 \: k\Omega\$ resistive, then (i) the gain and phase at DC would be \$\small 0.6\$ and zero, respectively (i.e. same as your results); and (ii) the gain and phase at \$\small 1590 \:\small Hz\$ would be \$\small 0.29\$ and \$\small -79^o\$, which compares with your measurement of \$\small 0.31\$ and \$\small -73^o\$.
Best Answer
No. Well, at least, not quite. The cutoff frequency only gives you pole behavior of your magnitude plot. I believe you're talking about the filter function where you basically perform a frequency sweep. However, depending on what the shape of the Bode plot, you can find a pattern from the magnitude plot. You haven't told us what filter this is nor its order so I am forced to give you an example.
If, for an example, you know that this amplifier circuit is a low pass filter, you can relate the frequency sweep and the cutoff frequency to the magnitude plot and phase plot.
Transfer function of the First Order Low Pass Filter:
\$ H(j\omega)=K\frac{1}{1+\frac{j\omega}{\omega_c}}\$
Magnitude Plot (and skipping some algebra):
\$M_{v-dB}=-10\log({1+\frac{\omega^2}{\omega_c^2}})\$
Lastly, the phase plot of the of the magnitude plot is:
\$\phi=-\tan^{-1}({\frac{\omega}{\omega_c}})\$
The attenuation can be determined by your typical \$V_{in}/V_{out}\$ equation in dB.
Hence,
\$K=(\frac{V_{in}}{V_{out}})_{dB}=10^{dB/20}\$.