Large pressure fluctuations are normal for nanoscale simulations of aqueous solutions (and other condensed systems without a gas phase).
Pressure fluctuations are usually enormous in simulations of aqueous solutions because water is nearly incompressible.
Let's say your system is a cube of water $(2~\mathrm{nm})^3$. Smaller systems will have larger fluctuations. At equilibrium, the variance of the pressure (its fluctuation) is
$$\tag{1}\sigma_P^2 = -kT \left( \frac{\partial P}{\partial V} \right)_S = \frac{kT}{\kappa_S V},$$
where $\kappa_S$ is the isentropic compressibility.
The isentropic compressibility is related to the density $\rho$ and speed of sound ($c$) by $\kappa_S = 1/(\rho c^2)$, so
$$\tag{2}\sigma_P^2 = \frac{\rho c^2 k T}{V}.$$
The standard deviation of the pressure should then be
$$\tag{3}\sigma_P = \sqrt{\frac{\rho c^2 k T}{V}}.$$
Taking the values for water near standard conditions $\rho=997~\mathrm{kg/m^3}, c=1500~\mathrm{m/s}, T=298.15~\mathrm{K}$ and using GNU units,
units 'sqrt(997*kg/m^3*(1500*m/s)^2*k*298*K/((2*nm)^3))' 'bar'
I get $\sigma_P \approx 340~\mathrm{bar}$. So the fluctuation of $\pm500~\mathrm{bar}$ is pretty reasonable.
You say you get a mean pressure of $-60~\mathrm{bar}$. What is the uncertainty in that calculation? The uncertainty of the mean pressure can be estimated by
$\delta P = \sigma_P/\sqrt{N}$, where $N$ is the number of independent data points for the pressure. So the number of independent (uncorrelated) data points you need to get to get the uncertainty $\delta P < 60~\mathrm{bar}$ is given by $N=\sigma_P^2/\delta P^2 \approx 30$. How long was your simulation? What is the autocorrelation time of the pressure (which determines the number of independent pressure values you have).
My guess is that your results are statistically consistent with a mean pressure of $1~\mathrm{bar}$.
Finally, for most molecular interactions in water, pressure–volume work ($p~\mathrm{d}V$) is negligible, so you don't have to worry much about the exact pressure as long as you don't have any weird vacuum bubbles that sometimes appear in simulations.
Aside: You might say, I'm using a barostat, shouldn't it force the pressure be constant? First, you have to realize that the instantaneous pressure is not well-defined and there isn't a unique way to calculate it. Some barostats might make some measures of instantaneous pressure constant, but this isn't true for the barostat I usually use. The likewise is true for thermostats. Instantaneous temperature is usually calculated from the average kinetic energy of a single simulation frame, which doesn't necessarily correspond to the thermodynamic temperature. Consider that a system with no momentum degrees of freedom (and no kinetic energy) can still have a thermodynamic temperature (e.g. the Ising model).