It depends on _where_ on Mars you toss the coin, and _how high_ you toss it.

In a rotating frame of reference, an object in motion appears to be affected by a pair of fictitious forces - the centrifugal force, and the Coriolis force. Their magnitude is given by

$$\mathbf{\vec{F_{centrifugal}}}=m\mathbf{\vec\omega\times(\vec\omega\times\vec{r})}\\
\mathbf{\vec{F_{Coriolis}}}=-2m\mathbf{\vec\Omega}\times\mathbf{\vec{v}}$$

The question is - when are these forces sufficient to move the coin "away from your hand" - in other words, for what initial velocity $v$ is the total displacement of the coin greater than 10 cm (as a rough estimate of what "back in your hand" might look like; obviously you can change the numbers).

The centrifugal force is always perpendicular to the axis of rotation, and is independent of the velocity of the particle (coin); this means that it will not affect the path of the coin at either the equator or the pole, but will displace it (towards the equator) at points in between. The Coriolis force is strongest at the equator, becoming zero at the pole; it is a function of the velocity of the coin.

To do the calculation properly, we have to compute the total displacement as a function of initial velocity $v_0$ and latitude $\theta$; then we can determine the "worst case" position and perhaps reach a conclusion.

As a simplifying assumption, we assume the change in height is sufficiently small that we ignore changes in the force of gravity; we also ignore all atmospheric drag (in particular, the wind; if the opening scene of "The Martian" were to be believed, it can get pretty windy on the Red Planet.) Finally we will assume that any horizontal velocity will be small - we ignore it for calculating the Coriolis force, but integrate it to obtain the displacement.

The vertical velocity is given by

$$v = v_0 - g\cdot t$$

and the total time taken is $t_t=\frac{2v_0}{g}$. At any moment, the Coriolis acceleration is

$$a_C=2\mathbf{\Omega}~v\cos\theta$$

Integrating once, we get

$$v_h = \int a\cdot dt \\
= 2\mathbf{\Omega}\cos\theta\int_0^t(v_0-gt)dt\\
= 2\mathbf{\Omega}\cos\theta\left(v_0 t-\frac12 gt^2\right)$$

And for the displacement

$$x_h = \int v_h dt \\
=  2\mathbf{\Omega}\cos\theta\int_0^t \left(v_0 t-\frac12 gt^2\right)dt\\
= 2\mathbf{\Omega}\cos\theta \left(\frac12 v_0 t^2-\frac16 gt^3\right)$$

Substituting for $t = \frac{2v_0}{g}$ we get

$$x_h = 2\mathbf{\Omega}\cos\theta v_0^3\left(\frac{4}{g^2} - \frac{4}{3 g^2}\right)\\
= \frac{16\mathbf{\Omega}\cos\theta v_0^3}{3g^2}$$

The sidereal day of Mars is 24 hours, 37 minutes and 22 seconds - so $\Omega = 7.088\cdot 10^{-5}/s$ and the acceleration of gravity $g = 3.71 m/s^2$. Plugging these values into the above equation, we find $x_h = 2.75\cdot 10^{-5}v_0^3 m$, where velocity is in m/s. From this it follows that you would have to toss the coin with an initial velocity of about 15 m/s for the Coriolis effect to be sufficient to deflect the coin by 10 cm before it comes back down. 

On Earth, such a toss would result in a coin that flies for about 3 seconds, reaching a height of about 11 m. It is conceivable that someone could toss a coin that high - but I've never seen it.

Now let's just see whether we need to worry about the effect of the centrifugal force. As I mentioned above,  it is proportional to the distance from the axis of rotation, and we need the component parallel to the surface. That gives us a dependance on latitude:

$$a_c = \mathbf{\Omega^2} R\cos\theta\sin\theta $$

To get an idea of the magnitude, let's calculate the displacement at 45°, where the centrifugal effect will be greatest. Since the acceleration is constant, we get the displacement directly from integrating twice:

$$x_c = \frac12 a t^2 = \frac12 \mathbf{\Omega^2} R\cos\theta\sin\theta \left(\frac{2v_0}{g}\right)^2$$

The dependence is on $v_0^2$ instead of $v_0^3$ since there is no velocity dependence of the centrifugal force: the harder you toss the coin, the less it matters (relatively speaking). At 15 m/s, I get

$$x_c = 0.28 m$$

So if you toss the coin at 45° latitude at a velocity of 15 m/s, it will not return to your hand - not because of the Coriolis effect (which would displace the coin in the East-West direction by a few cm) but because of the centrifugal force that moves the coin towards the equator. It follows that you could use a lower toss and still miss your hand - if we want the result to be 10 cm, velocity can come down by $\sqrt{2.8}$ to about 9 m/s - which is powerful but not crazy (corresponds to less than 5 m vertical toss on Earth).

That's a long winded way to say "yes, on Mars, at 45 degrees latitude, a coin tossed vertically with sufficient strength will miss your hand on the way down".