I concur with Avun Jahei's answer, but feel that it's worth quantifying the minimum non-catastrophic close approach distance.
This object has a mass comparable to Mars - 6.9 x 1023 kg
Earth's moon has a mass of 7.35 x 1022 kg and orbits at an average distance of 3.8 x 108 m.
For the purposes of calculating effects, let's assume that the rogue planet is exactly 10 times more massive than Earth's moon.
If the rogue planet passed by travelling on a trajectory perpendicular to the ecliptic and just outside lunar orbital distance then it would not be catastrophic for Earth's orbit - one never-repeated pass would be insufficient to have much effect - but it would be disastrous for Earth civilisation - massive tidal waves and energetic disruption to weather patterns, possible perturbation of the moon's orbit and so on. Quite apart from being freakishly unlikely that an approach would be that close - as Douglas Adams famously said, "Space is big" - this is probably not what you are looking for.
Fortunately, gravity works on an inverse square relationship. If we say that the rogue passed by with closest approach being 10 x lunar orbit away then it's tidal effects experienced on Earth will only be one-tenth that of the moon. The object is 10 times more massive, but the gravitational attraction is 100 times weaker at that distance (about 12-13 light-seconds). These effects would be somewhat noticeable, for example unusually high/low tides, but not catastrophic.
Increase the distance of closest approach by another order of magnitude and make closest approach about two light minutes away - still freakishly close for an interstellar object - and the effects from a single close encounter will be 0.1% as strong as lunar tidal effects. Scientists will be able to measure the effect and amateur telescopes will get a good look at it, but there will be no perceptible difference for people or other lifeforms on Earth.