Let's make this question a bit more operationally meaningful by asking if you can change the state of Pluto by choosing to do something here. As mentioned in the other answers, Pluto would feel the same force due to your mass even if you didn't exist, because the matter you consist of would be present on Earth anyway. However, you can still chose to move in a certain way and one can then consider the effect such a choice has on Pluto.
If you move then the distance between you an Pluto changes, if the distance changes from $d$ to $d+u$, the force will change. If $F(r)$ is the magnitude of the force exerted by you on Pluto, then we have:
$$\begin{split}
F(d+u) - F(d) &= G M_{\text{Pluto}}M_{\text{Antonio}}\left[\frac{1}{(d+u)^2}-\frac{1}{d^2}\right]\\&\approx G M_{\text{Pluto}}M_{\text{Antonio}}\left(-2\frac{u}{d^3}+3\frac{u^2}{d^4}\right)
\end{split}
$$
As pointed out in the comments by Dan and SchighSchagh, we must also take into account that the Earth moves in the opposite direction (actually, it's only part of the Earth as it cannot be treated as a rigid object, but that doesn't matter here), the center of mass doesn't change while the change in the force exerted on Pluto due to all the changes caused by the jump is to first order in $u$ due to the change in the center of mass. So, as pointed out by SchighSchagh, there is no net first order contribution.
The leading effect on Pluto is thus due to the second order term. The contribution due to the recoil of the Earth can then be ignored because the displacement squared times the mass for the Earth is now suppressed by the mass ratio of Antonio and the Earth relative to Antonio's contribution. We thus have:
$$F(d+u) - F(d) \approx 3G M_{\text{Pluto}}M_{\text{Antonio}}\frac{u^2}{d^4}
$$
To be precise we need to take into account that the change in the force experienced by Pluto now is due to the value of $u$ about 4.5 hours ago, so we need to use the so-called "retarded" value of the variable. Suppose then that Pluto will be overhead in 4.5 hours from now and you jump up to a height of $h$. The variable $u$ as a function of time will then be given by:
$$u(t) = -\sqrt{2 g h} t + \frac{1}{2}g t^2$$
for $0\leq t \leq \frac{2\sqrt{2 g h}}{g}$
The component of Pluto's momentum in the direction away from Earth will thus increase due to the jump by an amount of:
$$\Delta P_{\text{Pluto}}=\frac{3G M_{\text{Pluto}}M_{\text{Antonio}}}{d^4}\int_{0}^{\frac{2\sqrt{2 g h}}{g}}\left(\sqrt{2 g h} t - \frac{1}{2}g t^2\right)^2dt=\frac{4G M_{\text{Pluto}}M_{\text{Antonio}}}{15\sqrt{g}d^4}(2h)^{\frac{5}{2}}$$
Putting in the numbers here yields:
$$\Delta P_{\text{Pluto}}=7.9\times10^{-39}\frac{M_{\text{Antonio}}}{60\text{ kg}}\left(\frac{h}{\text{meter}}\right)^{\frac{5}{2}}\text{ Ns}$$
So, it seems that there is very small but real physical effect on Pluto. However when an extremely small amount of momentum is transferred, the physical state of the system may actually not change at all. This is due to the fact that the momentum of a system doesn't have a precise value due to quantum mechanics. A good example where you can see this effect at work is in certain variants of the double slit experiment where photons going through the slits will be bounced off mirrors before they hit the screen. If the imparted momentum on the mirrors would change the physical state of the mirror or the rest of the universe, then the interference pattern would vanish, because the information about which path the photon took will in principle exist. But in such experiments the interference pattern remains visible, which is experimental proof that the physical state of the rest of the universe actually does not change despite the momentum transfer.
To see if this effect is relevant, one has to give an approximate quantum mechanical description of Pluto's center of mass motion. Obviously, if Pluto were in some momentum eigenstate then the small momentum change would cause it's physical state to change, but obviously Pluto is not in such a state. A good approximation is obtained as follows. Pluto is not an isolated object, it receives energy from the Sun, its surface is at approximately 30 K. So, we can model it by assuming that all the degrees of freedom of Pluto are in a thermal bath at 30 K, and one of these degrees of freedom is its center of mass. What then happens is that due to the interactions with the thermal bath, the quantum mechanical uncertainty of the momentum of the center of mass is limited to about:
$$\Delta P_\text{QM}\approx \sqrt{M_{\text{Pluto}}k T}= 2.3 \text{ Ns}$$
So, the center of mass can be pictured as being described by an unknown wavefunction that in momentum space has a typical width of a few newton seconds. Since this is much larger than the transferred momentum, the state it would be in had you not jumped and the state it is in due to the jump have an overlap that is almost identical to 1. For the state to have changed unambiguously, the overlap between the two states should be zero. In terms of probabilities, one can say that Pluto will fail to detect whether or not you jumped with a probability of almost 1.