Why is Shamir Secret Sharing not secure against active adversaries out-of-the-box? This linear secret sharing scheme allows us to share a secret between n parties, such that only an honest majority can reconstruct it. 

I understand that – because I do not allow the user to certify the authenticity of the shares nor the value they reconstruct (using for instance VSS or some kind of homomorphic MAC) – they are secure only on the semi-honest case. However, in my mind that problem is more related to integrity, because that even if the adversary sends an incorrect share, the output is unpredictable, and as far as my understanding goes, indistinguishable from randomness. 

If that is the case, is there any way an active adversary can learn something by deviating from the protocol i.e., information about the private inputs of the honest parties? If that is so, what would it be?
Here is an active attack on the privacy of out-of-the-box SSS.  For this attack, we'll assume that the attacker (without a valid share) is allowed to participate (with $T-1$ friends with honest key shares), jointly use the protocol to recover a 'shared secret' (which might not be the real shared secret); we'll assume that this shared secret recovery process is done by some MPC-secure method so that the attacker learns nothing other than the final value.  We also assume that everyone's $x$ coordinate is public.

First off, the attacker picks a value $x_a$ which is an $x$ coordinate which is not the $x$-coordinate shared by anyone else.  If the dealer were to generate a share with that $x_a$ value, it would generate a $y$-coordinate that we'll call $y_a$; obviously, the attacker does not know this $y_a$ value.

Then, the attacker then gets a set of $T-1$ friends, and have them recover a shared secret value using their shares (and the attacker will contribute the value $(x_a, 0)$; (actually, a nonzero value for the $y$ coordinate would work, but make the explanation slightly more involved); they'll go through the SSS recovery process, and obtain a value which we'll denote as $S_0$.

If we call the real shared secret $S$, then what the attacker can do is derive a value $T_0$ (which depends on the $x$-coordinates of him and his friends), and derive the linear equation:

$$S = T_0 y_a + S_0$$

This is a linear equation in two unknowns ($S, y_a$), and $T_0 \ne 0$, and so the attacker cannot derive any information about $s$ from it.  This can also be derived from the security properties of SSS; the attacker and his $T-1$ friends only have $T-1$ valid shares, and so of course they cannot recover $S$.

What the attacker then does is get together a different set of $T-1$ friends (there can be some overlap, but there must be at least one difference), and do the same thing again (with the attacker using the same $(x_a, 0)$ share.  They do the same thing, deriving a 'shared secret' value $S_1$, and the attacker ends up with a second linear equation:

$$S = T_1 y_a + S_1$$

Now, if $T_0 \ne T_1$ (which is likely if the two sets of friends are distinct), the attacker has two linear equations in two unknowns; it's trivial to solve for $S$.