I'm travelling near the speed of light. Do I need to brake before using a super-massive black hole to turn around? I'm taking one of those new fusion drives for a trip to nearby supermassive black hole.  At a comfortable 1 G, it'll take me about 7 months of proper time to accelerate to 95% the speed of light.  I'm wondering, though, if I need to decelerate when I use the star's gravity to turn around and come home.  My first thought was that whipping around a star at 0.95c would create forces that would liquefy me and my crew.
My second thought was: this is a geodesic; even though we're completely reversing direction in a few microseconds, it will be just like floating in space.
Which is right? Do I need to brake for such a maneuver?
 A: 
My first thought was that whipping around a star at 0.95c would create forces that would liquefy me and my crew.

This is not correct, but there is a different problem that will likely prove fatal anyway.

My second thought was: this is a geodesic; even though we're turning around 180 degrees in a few microseconds, it will be just like floating in space.

This is closer to correct, but it has two serious problems: first is the size of the orbit required, and the second is the tidal forces involved.
Regarding the size of the orbit: in order to make a turn around at 0.95 c will require an orbit which passes exceptionally close to the central object.  For an object as large as a star you will not orbit outside the object but you will instead crash into the star itself.  The only objects which are sufficiently compact for you to pass that close are black holes.  This will require knowledge of your orbit and the black hole's position and motion to extreme precision.  Slightly too far away and your trajectory will barely deflect, and slightly too close and you will cross the event horizon.
Regarding the tidal forces: tidal forces are roughly proportional to $1/r^3$.  As you indicated above, for a point particle traveling along a geodesic is quite peaceful.  However, your ship and crew are not point particles.  The difference in forces will be enormous.  The parts of the ship on the inside of the turn will be making the 180 degree turn, while the parts of the ship on the outside of the turn will be barely deflected and continue on at nearly 0.95 c in nearly a straight line.
The only possible solution would be to find a supermassive black hole to turn around. That will reduce both of the above problems. (Or slow down)
A: 
Do I need to brake before using a star to turn around?

What kind of star? If the star is sufficiently compact, like neutron star or black hole, then no, you don't need to brake, and you'll be in free-fall the whole time. (Well, you might get fried by whatever plasma and electromagnetic fields are being slung around by the neutron star / black hole, but let's just consider gravitational effects here. No accretion disk, no astrophysical jets.)
As Dale's answer explained,  if the black hole is large enough, then tidal effects will be negligible, so the ride will be comfortable — if remaining weightless qualifies as "comfortable." A supermassive black hole should work quite nicely.
To make this a little more quantitative, first consider the simplest scenario: an ideal non-rotating black hole. Let $R$ denote the Schwarzschild radius, which is where the event horizon is. Light can orbit this black hole; the simplest orbit that light can have is a circular orbit, which necessarily has a radius of $3R/2$. And of course, if light can orbit the black hole, then the black hole can also be used to slingshot light. And if it can slingshot light, then it can certainly also slingshot something moving at 95\% of the spin of light. 
You'll need to tune the approach carefully, though, because slight changes in your approach to the black hole can make surprisingly big differences in the outcome. But with careful tuning, you can reverse your direction by slingshotting around a black hole.
Of course, real black holes are most likely spinning very rapidly. In that case, the frame-dragging effect makes life much more interesting. Yes, you can still slingshot around a rapidly-spinning black hole, but you'll need to do some very careful calculations to make sure you end up where you want to, because geodesics in the neighborhood of a Kerr (spinning) black hole can be crazy. 
For reference, first consider circular equatorial orbits for something moving at the speed of light. For a non-rotating black hole, the orbit must have radius $3R/2$, as mentioned above. The Schwarzschild radius in that case is $R=2M$, where $M$ is the black hole's mass (in natural units where Newton's constant and the speed of light are both equal to $1$), so the radius of the orbit can also be written $r=3M$. For a Kerr black hole that's spinning as fast as a black hole can possibly spin, the radius is either $r=M$ or $r=4M$ (equation 4.49 in "Geodesics of the Kerr metric", http://www.roma1.infn.it/teongrav/leonardo/bh/bhcap4.pdf), depending on whether the orbit is prograde or retrograde. This extreme difference between the photon-orbit radii in the prograde and retrograde cases is a testament to just how large the frame-dragging effects can be.
Those radii are for circular obrits moving at the speed of light, but the fact that such orbits exist implies that something moving at 95\% of the speed of light can use the black hole as a slingshot, as long as the distance of closest approach is suitably increased. And again, we're talking about geodesics here, so you'll remain comfortably(?) in free-fall the whole time, if the black hole is large enough to avoid uncomfortable tidal effects.
That's for equatorial geodesics. For non-equatorial geodesics, things get really wonky. For example, figure 4.25 in O'Neill's book The Geometry of Kerr Black Holes shows a "timelike flyby polar orbit that meets the Axis four times." The interactive website https://duetosymmetry.com/tool/kerr-circular-photon-orbits lets you construct examples of non-equatorial photon orbits, which illustrates just how careful you'll need to be with your calculations before attempting to use a spinning black hole as a slingshot (unless you're one of those people who thinks the journey is more important than the destination). But if it takes you $7$ months of proper time to accelerate to the desired speed, then I suppose you'll have plenty of free time to double-check those calculations.
Related:
https://space.stackexchange.com/q/530
https://space.stackexchange.com/q/1911
