IMO, the scenario is closer to your (a). I'll sketch an explanation of the duality between $H^1(E,\mathbf{Z}_l)$ and the dual to the Tate module. We have $H^1(E,\mathbf{Z}_l)=\text{Hom}(\pi_1(E),\mathbf{Z}_l)$,
where that $\pi_1$ means etale fundamental group with base point the origin $O$ of $E$. Thus the isomorphism we really want is between $\pi_1(E)\otimes\mathbf{Z}_l$ and $T_\ell(E)$.
What is $\pi_1(E)$? In the topology world, we'd consider the universal cover $f\colon E'\rightarrow E$ and take $\pi_1(E)$ to be its group of deck transformations. Then $\pi_1(E)$ has an obvious action on $f^{-1}(O)$. If $E$ is the complex manifold $\mathbf{C}/L$ for a lattice $L$, this is just the natural isomorphism $\pi_1(E)\cong L$.
But in the algebraic geometry world, there is no universal cover in the category of varieties, so the notion of universal cover is replaced with the projective system $E_i\to E$ of etale covers of $E$. Then $\pi_1(E)$ is the projective limit of the automorphism groups of $E_i$ over $E$.
One nice thing about $E$ being an elliptic curve is that any etale cover $E'\rightarrow E$ must also be an elliptic curve (once you choose an origin on it, anyway); if $E\rightarrow E'$ is the dual map then the composition $E\rightarrow E'\rightarrow E$ is multiplication by an integer. So it's sufficient to only consider those covers of $E$ which are just multiplication by an integer. Since it's $\pi_1(E)\otimes\mathbf{Z}_l$ we're interested in, it's enough to consider the isogenies of $E$ given by multiplication by $l^n$.
What are the deck transformations of the maps $l^n\colon E\rightarrow E$? Up to an automorphism of $E$, they're simply translations by $l^n$-division points. And now we see the relationship to the Tate module: A compatible system of deck transformations of these covers is the exact same thing as a compatible system of $l^n$-division points. Thus we get the desired isomorphism. Naturally, it's Galois compatible!
In the end, we see that torsion points were tucked away in the construction of the etale cohomology groups, so it wasn't exactly a coincidence. Hope this helps.
Re the edit: I believe your best bet is to work locally. First of all, you didn't mention which Galois representation you wanted exactly; let's say you want the representation on $H^i$ of your variety for a given $i$. Let's assume this space has dimension $d$.
Step 1. For each prime $p$ at which your variety $V$ has good reduction, you can compute the local zeta function of $V/\mathbf{F}_p$ by counting points on $V(\mathbf{F}_{p^n})$ for $n\geq 0$. In this way you can compute the action of the $p^n$th power Frobenius on $H^i(V\otimes\overline{\mathbf{F}}_p,\mathbf{F}_5)$ for various primes $p$.
Step 2. Do this enough so that you can gather up information on the statistics of how often the Frobenius at $p$ lands in each conjugacy class in the group $\text{GL}_d(\mathbf{F}_5)$. In this way you could guess the conjugacy class of the image of Galois inside $\text{GL}_d(\mathbf{F}_5)$.
Step 3. Now your job is to find a table of number fields $F$ whose splitting field has Galois group equal to the group you found in the previous step. I found a table here: http://hobbes.la.asu.edu/NFDB/. You already know which primes ramify in $K$ -- these are at worst the primes of bad reduction of $V$ together with 5 -- and you can distinguish your $F$ from the other number fields by the splitting behavior your found in Step 1. Then $K$ is the splitting field of $F$.
A caveat: Step 1 may well take you a very long time, because unless your variety has some special structure or symmetry to it, counting points on $V$ is Hard.
Another caveat: Step 3 might be impossible if $d$ is large. If $d$ is 2 then perhaps you're ok, because there might be a degree 8 number field $F$ whose splitting field has Galois group $\text{GL}_2(\mathbf{F}_5)$. If $d$ is large you might be out of luck here.
You are free not to accept this answer because of the above caveats but I really do think you've asked a hell of a tough question here!
This is not an answer, it's just a note that some of the more standard examples of functors don't seem to work because they're not fully faithful. This was originally in the comments but it clogged them up so I deleted them and moved them here.
The observation I want to make is simply that if $k=\mathbb{Q}_p$ then I think that none of the following work in general: $K=\mathbb{Q}_\ell$ and the functor is $\ell$-adic etale cohomology; $K=\mathbb{Q}_p$ and the functor is $p$-adic etale cohomology; and finally $K=\mathbb{Q}_p$, the target category is $K$-vector spaces equipped with a filtration, a Frobenius and a monodromy operator in the usual $p$-adic Hodge theory waym and the functor is $D_{pst}$.
The reason none of these work is the following. Fix an elliptic curve over $\mathbb{F}_p$ with $a_p=0$ (assume such a curve exists). Then there are uncountably many lifts of this curve to $\mathbb{Q}_p$, so there are two non-isogenous lifts. The motives attached to the $h^1$ of these curves will, I imagine, have no non-zero maps between them. However the functors described above are all controlled by the special fibre so produce isomorphic linear algebra objects. In the $D_{pst}$ case what is happening is that the Frobenius has trace zero and the $Fil^1$ can't be either of the eigenspaces so it must be another line; however all other lines give isomorphic filtered $\phi$-modules.
The reason this doesn't contradict the Hodge conjecture is that you can take coefficient field equal to $\mathbb{Q}$ in that setting, which is much smaller, and then the filtration (which is defined once you tensor up to $\mathbb{C}$ gives you far more information.
Best Answer
Yes. In fact what Artin proves in SGA4 exp XI thm 4.4 is that étale cohomology and singular cohomology agree for smooth schemes over $\mathbb{C}$ with finite coefficients. The statement you want will follow from this by taking inverse limits to get to $\mathbb{Z}_\ell$ and then extending scalars to $\mathbb{Q}_\ell$. If you don't feel like looking at SGA, you can find treatments of this in the books by Freitag-Kiehl, Milne,…
Added (in response to comment). The isomorphism $H_{et}^*(X_{\bar K}, \mathbb{Q}_\ell)\cong H_{et}^*(X_{\mathbb{C}}, \mathbb{Q}_\ell)$ follows from the smooth base change theorem (cf. Milne, Etale cohomology, p 231 cor 4.3).