$\mathrm{H}^1(G,\mathbb{Z}/2) \cong \hom(G,\mathbb{Z}/2)$ consists precisely of first Stiefel–Whitney classes.

So your question is to decide whether there is a finite group with an element in $\mathrm{H}^2(G,\mathbb{Z}/2)$, i.e. with a double cover, which is not a second Stiefel–Whitney class, i.e. which is not pulled back from the Spin double cover of an oriented real representation. (What about $w_1$s? Well, first, if $A$ and $B$ are line-bundles, then $w_2(A\oplus B) = w_1(A)w_1(B)$. Second, for $V$ arbitrary, $V \oplus 3\det(V)$ is orientable and $w_2(V \oplus 3\det(V)) = w_2(V)$.)

I don't know the answer, but I can answer in the negative the version where $G$ is a Lie group. Namely, consider the double cover $\mathrm{SO}(8) \to \mathrm{PSO}(8)$. I claim that this double cover is not pulled back from any Spin double cover of any representation $V$ of $G = \mathrm{PSO}(8)$. Equivalently, I claim that the class in $\mathrm{H}^2(B\mathrm{PSO}(8), \mathbb Z/2)$ classifying this double cover is not a Stiefel–Whitney class.

It suffices to restrict the map $\mathrm{PSO}(8) \to \mathrm{SO}(V)$ to the maximal torus $T \subset \mathrm{PSO}(8)$, where I can calculate with weights. The root lattice $\Lambda$ of $T$ is the sublattice $\Lambda \subset \mathbb Z^4$ consisting of vectors with even dot product with $(1,1,1,1)$. The roots are $(\pm1,\pm1,0,0)$, $(\pm1,0,\pm1,0)$, $(\pm1,0,0,\pm1)$, $(0,\pm1,\pm1,0)$, $(0,\pm1,0,\pm1)$, and $(0,0,\pm1,\pm1)$. The weights of the vector representation of $\mathrm{SO}(8)$ are $(\pm1,0,0,0)$, $(0,\pm1,0,0)$, $(0,0,\pm1,0)$, and $(0,0,0,\pm1)$. The Weyl group is $W = 2^3{:}4!$, where the $2^3$ normal subgroup acts as reflections in an even number of coordinates, and $4! = S_4$ acts as permutations.

I will use only that $V|_T$ is a sum of Weyl-invariant representations of $T$. I.e. I claim for any Weyl-invarnat representation $T \to \mathrm{SO}(V)$, the Spin double cover of $\mathrm{SO}(V)$ trivializes when restricted to $T$.

An irreducible representation $V$ of $T.W$, i.e. a Weyl-invariant irrep of $T$, consists of a single Weyl orbit through some vector $\lambda \in \Lambda$. Note that $-1 \in W$. It follows that $V$ is the underlying real representation of a complex representation $U$ of $T$. (Emphasis: this holds over $T$, but not over $T.W$ or $\mathrm{PSO}(8)$.) For instance, you can take $U$ to consist of the positive weights appearing in $V$. (The positive weights are those for which the first nonzero entry is positive.)

It is a general fact that if $V$ is the underlying real representation of a complex representation $U$, then $w_2(V) = c_1(U) \mod 2$. But $c_1(U)$ is exactly the sum of weights appearing in $U$. And my claim that $w_2(V) = 0$ is equivalent to claiming that $\frac12 c_1(U) \in \Lambda$, which is to say I need to show that $\frac12 c_1(U) \in \mathbb Z^4$ and that $\langle \frac12 c_1(U), (1,1,1,1) \rangle \in 2\mathbb Z$.

I'll let the highest weight of $V$ be $\lambda = (a,b,c,d) \in \Lambda$, so that $a+b+c+d \in 2\mathbb Z$ (and remember that my $V$ is simply the Weyl-orbit through that highest weight). The positive Weyl chamber is $a \geq b \geq c \geq d \geq -c$, so that's where $\lambda$ lives. The "reflection" outer automorphism sends $(a,b,c,d) \mapsto (a,b,c,-d)$, so I can assume $d \geq 0$.

I'll casebash the calculation.

Case 1: $a>0$, $b=c=d=0$. The weights of $V$ are $(\pm a, 0,0,0), (0, \pm a,0,0)$, $(0,0,\pm a,0)$, $(0,0,0,\pm a)$, so the weights of $U$ are $(a,0,0,0)$, $(0,a,0,0)$, $(0,0,a,0)$, $(0,0,0,a)$, and $c_1(U) = (a,a,a,a)$. Note that $a \in 2\mathbb Z$ by assumption, and so $\frac12 c_1(U) \in \mathbb Z^4$. Furthermore $\frac12 \langle c_1(U),(1,1,1,1)\rangle = 2a$ is even (in fact divisible by $4$).

Case 2: $a \geq b >0$, $c = d = 0$. If $a=b$, then the weights of $U$ are $(a, \pm a, 0,0)$, $(a, 0, \pm a, 0)$, $(a, 0,0\pm a, )$, $(0,a,\pm a,0)$, $(0,a,0,\pm a)$, $(0,0,a,\pm a)$ and $\frac12 c_1(U) = (3a,2a,a,0) \in \mathbb Z^4$ and $3+2+1 = 6$ is even. If $a > b$, then $\frac12 c_1(U) = (3(a+b),2(a+b),a+b,0)$. In either case, we get a term divisible by $6$ for each permutation of the $(a,b)$.

Case 3: $c>0$, $d=0$. The weights of $U$ come in four sets: $(+x, \pm y,\pm z,0)$, $(+x, \pm y, 0, \pm z)$, $(+x, 0, \pm y, \pm z)$, and $(0, +x, \pm y, \pm z)$, where $(x,y,z)$ ranges over the permutations of $(a,b,c)$. Then $\frac12 c_1(U)$ is a sum over permutations of $(a,b,c)$ of a term like $(6x, 2x, 0,0)$, manifestly integral and with integer dot-product with $(1,1,1,1)$.

Case 4: If $d>0$, then the weights of $U$ are the vectors of the form
$$ (+w, +x, +y, +z), (+w, +x, -y, -z), (+w, -x, +y, -z), (+w, -x, -y, +z)$$
where $(w,x,y,z)$ ranges over the permutations of $(a,b,c,d)$. Then $\frac12 c_1$ is a sum of terms like $(2w,0,0,0)$.

Note that the only case where $\frac12 c_1(U)$ came close to leaving $\mathbb Z^4$ was Case 1, where I needed that $a \in 2\mathbb Z$, and the only case where $\langle \frac12 c_1(U), (1,1,1,1)\rangle$ came close to leaving $2\mathbb Z$ was Case 2.

I remark that the triality outer automorphism relates $\lambda = (2,0,0,0)$ (Case 1) to $\lambda = (1,1,1,1)$ (Case 4). For the former, $\frac12 c_1(U) = (1,1,1,1)$, and for the latter $\frac12 c_1(U) = (2,0,0,0)$.

If you repeat the argument for a general PSO group, I think you will find that all real representations of $\mathrm{PSO}(n)$ are Spin (and so you get no nontrivial double covers from them) whenever $n$ is divisible by $8$.