No. Take PerfectGroup(245760,4) (this requires GAP 4.12 :-) ), generated by
(1,11,10,8,14)(2,4,12,15,16)(3,13,5,6,7)(17,25,28,21,23)(18,19,24,30,29)(20,27,22,26,31)(33,45,39,41,35)(34,47,42,36,46)(38,40,48,44,43),
(1,13,12,16,9)(2,7,5,6,14)(3,10,15,8,4)(17,21,23,27,30)(18,24,32,22,28)(19,20,29,25,31)(33,40,45,36,37)(34,38,48,39,41)(35,46,42,44,47),
(1,14,2)(3,12,16)(4,9,7)(5,15,8)(6,13,10)(17,31,18)(19,28,29)(20,25,23)(21,32,24)(22,30,26)(33,39,38)(34,43,47)(35,36,45)(37,42,48)(41,46,44)
It has a structure $(2^4\times 2^4\times 2^4):A_5$, where $2^4$ is the irreducible $A_5$-module of dimension 4 that does not come from the permutation representation. It has 21 minimal normal subgroups (all of type $2^4$). The solvable radical $(2^4)^3$ is composed from 21 classes of order 15 (which each generate one of the minimal normal subgroups) and 63 classes of order 60 (that each generate a normal subgroup of order 256), and of course the identity.
This group has trivial center (thus cyclic) but has no faithful irreducible representation.
This is the smallest possible example, and the only one of order 245760. (There are further ones in order 491520, not neccessarily just extensions with this one.)
Best Answer
No. A quasisimple group has a faithful irreducible representation if and only if the centre is cyclic. Since there are simple groups with non-cyclic Schur multiplier (e.g., $PSL_3(4)$, which has Schur multiplier $4\times 4\times 3$), there are no faithful irreducible representations of $4^2.PSL_3(4)$.
Edit: I should say, morally the answer is 'yes', because finite simple groups have cyclic Schur multiplier. The problem is for some small groups, for example $PSL_3(4)$, $Sz(8)$, $PSU_4(3)$, this isn't true, and also, as one comment just reminded me, the orthogonal groups $\Omega_{4n}^+(q)$ for $q$ odd and $n\geq 2$. But there are only finitely many such examples, apart from the orthogonal groups.
Edit 2: Just to clarify, the situation for quasisimple groups is that they have faithful, irreducible representations if and only if they have a cyclic centre. This is not difficult to see by inducing a faithful linear character for the centre. In general, this is not sufficient. This MathOverflow post gives information about the general case.