I've been learning about p-adic Hodge theory recently (I'm a beginner), and I've been wondering about the following question the past couple of weeks. Sorry for the long setup, it's mainly background; the question is down towards the bottom.

Background: For a field $k$, and a second field $K$ of characteristic 0, we can consider the $K$-linear category $Mot(k)_{K}$ of motives defined by smooth projective varieties over $k$ (pure motives) up to homological or numerical equivalence (let's assume homological equivalence equals numerical equivalence for each Weil cohomology we consider). Then this is a semi-simple abelian category (due to Jannsen), and there are realization functors to various categories of vector spaces extending the functor sending a variety $X$ to its cohomology groups (e.g. its etale cohomology groups $H^i(X,\mathbb{Q}_\ell)$ for $\ell \not= p$ in which case $K = \mathbb{Q}_\ell$).

In the case when $k = \mathbb{C}$ and we take $K = \mathbb{Q}$, we can consider the singular cohomology realization $Mot(\mathbb{C})_\mathbb{Q} \rightarrow {Vect}_{\mathbb{Q}}$ sending $X$ to $H_{sing}^*(X,\mathbb{Q})$. One way of stating the Hodge conjecture is that when we consider the extra structure of the Hodge decomposition on $H^*_{sing}(X,\mathbb{Q}) \otimes \mathbb{C}$ and therefore consider the enriched realization $Mot(\mathbb{C})_{\mathbb{Q}} \rightarrow {Hdge}_{\mathbb{Q}}$ to pure $\mathbb{Q}$-Hodge structures, this functor is fully faithful.

Similarly, if $k$ is a finite field or a number field, we can take $K = \mathbb{Q}_\ell$, we can consider the functor $Mot(k)_{\mathbb{Q}_{\ell}} \rightarrow GalRep(\mathbb{Q}_\ell)$ which sends $X$ to its $\ell$-adic cohomology $H^i(X_{\overline{k}},\mathbb{Q}_\ell)$ as an $\ell$-adic Galois representation. Then the Tate conjecture is that this functor is fully faithful. In the number field case, there are some other conjecturally fully faithful enriched realization functors discussed in Ch. 7 of Andre's book "Une Introduction Aux Motifs".

Motivation: Now in $p$-adic Hodge theory, we deal with varieties over a complete discrete valuation field (for example $k = \mathbb{Q}_p$), and study the extra structures which one obtains on the de Rham cohomology of such a variety from Hodge theory, crystalline cohomology, and etale cohomology. To me it seems that one is trying to endow the de Rham cohomology of a variety $X$ with as much "natural" structure as possible, at least enough to determine the Galois representation $H^i(X_{\overline{k}},\mathbb{Q}_p)$, but possibly one can add more. So I wonder the following:

Let $k$ be a finite extension of $\mathbb{Q}_p$. Is it reasonable to believe that there exist a characteristic-0 field $K$ and a fully faithful functor $Mot(k)_K \rightarrow \mathcal{C}$, where $\mathcal{C}$ is the category of $K$-vector spaces with some "natural" extra structure, for example that considered in $p$-adic Hodge theory? If not, why not? If so, is there a candidate for $\mathcal{C}$?

EDIT: I want to add an additional condition: that when one takes the forgetful functor $\mathcal{C} \rightarrow Vect(K)$, the composition $Mot(k)_K \rightarrow Vect(K)$ is the functor defined by some Weil cohomology (e.g., de Rham cohomology or $p$-adic etale cohomology).

My impression from the example given by user ulrich in the comment to

Should the etale cohomology of a smooth projective variety (over rationals) be semi-simple; why?

is that we CANNOT take $K = \mathbb{Q}_p$ and take the functor sending $X$ to $H^*(X_{\overline{k}},\mathbb{Q}_p)$ , because the the $p$-adic cohomology of $X$ might not be semi-simple, for example when $X$ is an elliptic curve with non-split multiplicative reduction. But perhaps there's another possible $\mathcal{C}$?

EDIT: As wccanard explained in a comment (now deleted), the example above doesn't actually show that the $p$-adic etale cohomology functor isn't fully faithful, since the subobject of $H^1(X,\mathbb{Q}_p)$ which is not split is not "motivic". However, there are other reasons why $p$-adic etale cohomology can't work as explained in wccarnard's answer below.

canonicallyisomorphic, so one can see the different varieties giving different Hodge filtrations. But this doesn't help you here, because $\mathcal{C}$ can't tell the canonical isomorphism from all of the non-canonical ones :-/ – user30035 Mar 19 '13 at 22:54