You cannot say this without telling which specific measurement device you use. What happens depends on the specific interaction of the measured system with the measurement device.
In your case of the free particle, you should note that there are no energy eigenstates in the literal sense, because the energy eigenstates are uniquely given by the common eigenstates of thee three momentum components, and these are plane waves. They are not square-integrable and thus belong not to the Hilbert space (in wave mechanics the space of square-integrable functions) but to the dual of the nuclear space, where position and momentum operators are defined, i.e., they are distributions.
True states, coming close to energy eigenstates are square-integrable wave functions which peak quite sharply in momentum space. You can, e.g., choose a Gaussian ##\tilde{\psi}_0(\vec{p}) \propto \exp[-(\vec{p}-\vec{p}_0)^2/(4 \sigma_p)]## as an initial state. Then the solution of the Schrödinger equation in momentum space simply is
$$\tilde{\psi}(t,\vec{p})=\exp \left (-\frac{\mathrm{i} \vec{p}^2}{2m \hbar} t \right) \tilde{\psi}_0(\vec{p}).$$
The position representation then follows by Fourier transformation
$$\psi(t,\vec{x})= \frac{1}{\sqrt{2 \pi \hbar}} \int_{\mathbb{R}^3} \mathrm{d}^3 p \tilde{\psi}(t,\vec{x}),$$
which is again a Gaussian, which now however is rather broad, and the position uncertainty increases with time.
Always the Heisenberg uncertainty relation holds,
$$\Delta x_j \Delta p_j \geq \hbar/2$$
for each position-vector and momentum component, ##j \in \{1,2,3\}##.
