Let $ G $ be a (non-abelian) simple finite group. Then the image of a degree $ d $ representation $ \pi: A \to U_d $ always lies in $ SU_d $.
Indeed this is true even if $ G $ is quasisimple.
What if $ G $ is almost simple? It seems obvious that this would no longer be true. For a reducible representation this is easy to show. The natural permutation representation $ \pi: S_5 \to U_5 $ has half matrices $ det=-1 $ thus is not contained in $ SU_5 $.
It seems almost ridiculous to ask this because I think a counterexample should be obvious but I can't think of one:
What is an example of a faithful degree $ d $ irreducible representation $ \pi: G \to U_d $ of an almost simple finite group $ G $ whose image is not in $ SU_d $?
The obvious first place to look is the faithful irreps of $ S_5 $. These all map into $ O_4, O_5 $ or $ O_6 $. However it's not obvious to me whether or not these map into $ SO_4,SO_5,SO_6 $.
Best Answer
Answer from Derek Holt given in the comments:
The conjugacy class of (1,2) in $ S_5 $ has character value $ 2 $. Since $ (1,2) $ is an involution all its eigenvalues are $ \pm 1 $. Since the four eigenvalues sum to $ 2 $ and are all $ \pm 1 $ they must be $ 1,1,1,-1 $ and thus the image of $ (1,2) $ in such a representation must have determinant $ -1 $, in other words it must be in $ O_4(\mathbb{R}) $ not in $ SO_4(\mathbb{R}) $.
Here is a concrete example. The matrices $ F_1,F_2',F_3' $ in
https://arxiv.org/abs/hep-th/9905212
generate a 4d irrep of $ A_5 $. Adding in the matrix $ CNOT=\overline{\zeta_8} F' $ (just $ F' $ from top of page 7 but without the phase) $$ CNOT:=\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} $$
generates a 4d irrep of $ S_5 $ where the det 1 subgroup is exactly $ A_5 $ and the rest of $ S_5 $ all has determinant $ -1 $.