Negative answer
Karpilovsky's result is not useful here. It deals with GL-conjugacy, not PGL-conjugacy.
We setup two projective representations of $G=A_4$ that have the same defining cocycle $\alpha$, whose $\alpha$-characters are different, but such that the representations are conjugate in the projective general linear group.
Notice that the $\alpha$-characters are not class functions, but that one could choose $\alpha$ so that $\chi_1$ becomes a class function.
The example
$ \newcommand{\ze}{\zeta_3} \newcommand{\zi}{\ze^{-1}} \newcommand{\vp}{\vphantom{\zi}} \newcommand{\m}[1]{\left[\begin{smallmatrix}\vp #1 \vp\end{smallmatrix}\right]} \newcommand{\PGL}{\operatorname{PGL}} $ Let $R$ be a commutative ring with identity and a primitive 3rd root of unity $\ze$.
Define $\rho_1: A_4 \to \PGL(2,R)$ by $\rho_1((12)(34)) = \m{ 0 & 1 \\ -1 & 0 }$ and $\rho_1((123) ) = \m{ \ze & 0 \\ -1 & \zi }$.
Define $\rho_2: A_4 \to \PGL(2,R)$ by $\rho_2((12)(34)) = \m{ 0 & 1 \\ -1 & 0 }$ and $\rho_2((123) ) = \m{ 0 & -\zi \\ 1 & -\ze }$.
Note that for $x=\left[\begin{smallmatrix}\ze & 1 \\ -1 & \ze \end{smallmatrix}\right]$ we get $\rho_1((12)(34))^x = \rho_2((12)(34)) \quad\text{and}\quad \rho_1((123))^x = \zi \cdot \rho_2((123))$ so that the projective representations are conjugate in $\PGL(2,R)$. Note that all matrices mentioned so far are invertible over $R$.
Explicitly we get the following values (the choice of signs defines the cocycle; we've chosen signs for $\rho_1$ and $\rho_2$ to give the same cocycle, $\alpha$):
$\tiny\begin{array}{r|c|ccc|ccccccccc} & () & (12)(34) & (13)(24) & (14)(23) & (123) & (132) & (124) & (142) & (134) & ( 143) & (234) & (243) \\ \hline \rho_1 & \m{ 1 & 0 \\ 0 & 1} & \m{ 0 & -1 \\ 1 & 0} & \m{ -\ze & -\zi \\ -\zi & \ze} & \m{ -\zi & \ze \\ \ze & \zi} & \m{ -\ze & 0 \\ 1 & -\zi} & \m{ -\zi & 0 \\ -1 & -\ze} & \m{ \ze & -1 \\ 0 & \zi} & \m{ \zi & 1 \\ 0 & \ze} & \m{ -1 & \zi \\ -\ze & 0} & \m{ 0 & -\zi \\ \ze & -1} & \m{ -1 & -\ze \\ \zi & 0} & \m{ 0 & \ze \\ -\zi & -1} \\ \rho_2 & \m{ 1 & 0 \\ 0 & 1} & \m{ 0 & -1 \\ 1 & 0} & \m{ \ze & -\zi \\ -\zi & -\ze} & \m{ -\zi & -\ze \\ -\ze & \zi} & \m{ 0 & \zi \\ -1 & \ze} & \m{ \zi & -1 \\ \ze & 0} & \m{ 0 & \ze \\ -1 & -\zi} & \m{ -\ze & -1 \\ \zi & 0} & \m{ 1 & -\ze \\ 0 & \zi} & \m{ 1 & \zi \\ 0 & \ze} & \m{ \ze & 0 \\ -\zi & 1} & \m{ \zi & 0 \\ \ze & 1} \end{array}$
$\chi_i(g) := \operatorname{tr}(\rho_i(g))$ has values:
$\scriptsize\begin{array}{r|c|ccc|ccccccccc} & () & (12)(34) & (13)(24) & (14)(23) & (123) & (132) & (124) & (142) & (134) & ( 143) & (234) & (243) \\ \hline \chi_1 & 2 & 0 & 0 & 0 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \chi_2 & 2 & 0 & 0 & 0 & \ze & \zi & -\zi & -\ze & -\ze & -\zi & -\zi & -\ze \end{array}$
Note that these characters are not equal, but the representations $\rho_i$ are conjugate in $\operatorname{PGL}(2,R)$. The lifts of these representations to $\operatorname{GL}(2,R)$ are non-isomorphic representations of $\operatorname{SL}(2,3)$.
For reference, $\alpha$ is:
$\tiny \begin{array}{r|r|rrr|rrrrrrrr} \alpha & () & (12)(34) & (13)(24) & (14)(23) & (123) & (132) & (124) & (142) & (134) & ( 143) & (234) & (243) \\ \hline () & 1&1&1&1&1&1&1&1&1&1&1&1\\ \hline (12)(34) & 1&-1&-1&1&1&-1&1&-1&-1&-1&1&1\\ (13)(24) & 1&1&-1&-1&1&-1&1&1&-1&1&-1&-1\\ (14)(23) & 1&-1&1&-1&-1&1&1&1&-1&-1&-1&1\\ \hline (123) & 1&1&1&-1&-1&1&1&1&1&-1&-1&1\\ (132) & 1&-1&-1&1&1&-1&1&1&-1&1&1&1\\ (124) & 1&1&-1&-1&-1&1&1&1&-1&1&-1&-1\\ (142) & 1&-1&-1&-1&1&-1&1&1&-1&-1&1&1\\ (134) & 1&1&-1&1&1&1&1&1&1&1&1&-1\\ (143) & 1&1&1&1&1&1&-1&1&1&1&1&-1\\ (234) & 1&-1&1&-1&-1&1&1&1&1&-1&1&1\\ (243) & 1&-1&1&1&1&1&-1&-1&1&-1&1&1 \end{array}$