I think the statement can fail in the case of elliptic curves of good reduction even when $l = p$. But then your comment on
Serre-Tate theory confused me for a little while! (A discussion with Jared Weinstein helped me clear it up.)
The post ended up getting long as a result; it's partly for my own reference.
Recently FC pointed out the following to me: if $E$ over $\mathbb Q_p$ is an elliptic curve of good supersingular reduction
and $p > 3$, then $T_p(E)$ (up to isomorphism) does not depend on $E$.
First, $a_p = 0$ by the Weil bounds. The $p$-adic Galois representation $\rho := T_p(E) \otimes \mathbb Q_p$ is crystalline
(due to good reduction) of Hodge-Tate weights 0, 1 with (crystalline) $a_p = 0$. In particular, it is non-ordinary so $\rho$
is (even residually) absolutely irreducible. It's a basic fact in $p$-adic Hodge theory that any 2-dim. absolutely
irreducible $G_{\mathbb Q_p}$-representation with distinct Hodge-Tate weights is uniquely determined by $a_p$. (This is
nowadays an exercise about weakly admissible filtered $\phi$-modules.) Finally, as $\rho$ is residually irreducible, it
follows that the lattice $T_p(E)$ is unique up to homothety (in particular, up to isomorphism).
Just as Tim notes, isogeny classes are countable. But there are uncountably many elliptic curves with good supersingular
reduction. (Fix a supersingular one over $\mathbb F_p$ and consider lifts of the $j$-invariant.)
I wonder how general a phenomenon this is that the Galois representation doesn't vary.
Kisin's result no longer seems to apply in this $l = p$ case (non-torsion coefficients).
Now, here is why there is no contradiction with Serre-Tate theory. First, it is true that any two of the elliptic curves
above have isomorphic $p$-divisible groups over $\mathbb Z_p$: the reason is that $T_p(E)$ (with the Galois action) precisely
tells us what $E[p^\infty]$ is over $\mathbb Q_p$, and that Tate's full-faithfulness theorem says that this determines
$E[p^\infty]$ over $\mathbb Z_p$. Besides, the special fibre $E/\mathbb F_p$ is uniquely determined up to isogeny,
because we know $a_p$. [There's even an isogeny over $\mathbb F_p$, e.g., because the Dieudonne module is the same (see
below), but there must be an elementary way to see that.]
Serre-Tate theory says the following (specialised to our situation). The category of elliptic curves over $\mathbb Z_p$ is
equivalent to the following category $C$: the objects consist of triples $(\mathcal G, \overline E, f : \mathcal G \times \mathbb
F_p \to \overline E[p^\infty])$, where $\mathcal G$ is a $p$-divisible group over $\mathbb Z_p$, $\overline E$ is an elliptic curve
over $\mathbb F_p$ and $f$ is an isomorphism of $p$-divisible groups over $\mathbb F_p$. Morphisms consist of a map of
$p$-divisible groups over $\mathbb Z_p$ and a map of elliptic curves over $\mathbb F_p$ that are compatible with the
isomorphisms in the special fibre. In this equivalence, an elliptic curve $E/\mathbb Z_p$ maps to the triple $(E[p^\infty],
E \times \mathbb F_p, can)$, where $can$ is the canonical isomorphism.
Suppose we have two elliptic curves $E_1$, $E_2$ over $\mathbb Z_p$ as above (i.e., supersingular reduction and $p >
3$). Then we know that $E_1[p^\infty] \cong E_2[p^\infty]$ and that $E_1 \times \mathbb F_p$ is isogenous to $E_2 \times
\mathbb F_p$. But in order to apply the theorem of Serre-Tate and deduce the existence of an isogeny $E_1 \to E_2$ we need to
know that we can choose a map $\alpha : E_1[p^\infty] \to E_2[p^\infty]$ of $p$-divisible groups over $\mathbb Z_p$ and a map
$\beta : E_1 \times \mathbb F_p \to E_2 \times \mathbb F_p$ of elliptic curves such that
$\alpha \times \mathbb F_p = \beta[p^\infty]$ as map of $p$-divisible groups $E_1[p^\infty] \to E_2[p^\infty]$ over $\mathbb F_p$.
The problem is that we have very little flexibility: there are only countably many homomorphisms $E_1 \times \mathbb F_p
\to E_2 \times \mathbb F_p$. But it's hard to lift a map of $p$-divisible groups over $\mathbb F_p$ to a map
of $p$-divisible groups over $\mathbb Z_p$:
The category of $p$-divisible groups over $\mathbb Z_p$ is equivalent, by results of Fontaine-Laffaille, to the
category of quadruples $(M,M^1,\phi,\phi^1)$, where $M$ is a finite free $\mathbb Z_p$-module, $M^1$ a $\mathbb Z_p$-direct
summand ("Hodge filtration"), and $\phi : M \to M$, $\phi^1 : M^1 \to M$ are $\mathbb Z_p$-linear maps such that $p\phi^1 = \phi|_{M^1}$ and $\phi(M) +
\phi^1(M^1) = M$. [In our case $M$ is of rank 2 and $M^1$ of rank 1.]
The Dieudonne module of the special fibre of the given $p$-divisible group is recovered as follows
(reference?): we need to give a $\mathbb Z_p[F,V]/(FV-p)$-module $D$ that is finite free as $\mathbb Z_p$-module. We take $D
:= M$, $F := \phi$, and $V := pF^{-1}$ (this is defined on $M[1/p]$ initially, but the final condition on $M$ forces it to
preserve $M$). In other words, one forgets the Hodge filtration. So lifting $\beta[p^\infty]$ to $\mathbb Z_p$ is difficult because a given map of Dieudonne
modules doesn't need to preserve the Hodge filtration. (The countable flexibility we have with $\beta$ is not enough to
move a given line to any other. Note that $M^1 \otimes \mathbb F_p$ is uniquely determined as the kernel of $\phi \otimes \mathbb F_p$; but apart from that $M^1$ is free to vary.)
[Finally, a small comment/question: I think Tate's result over finite fields assumes that $l$ is prime to the characteristic.
Milne-Waterhouse (1971) have a version for $l = p$ where $T_l$ is replaced by the Dieudonné module. In this case
$T_p E$ is just the first crystalline cohomology group. Is there a general crystalline analogue of the Tate conjecture?]
Added: I think the isogeny theorem can also fail in the case of good ordinary reduction, using the same argument.
Consider again elliptic curves over $\mathbb Q_p$, now of good ordinary reduction. Let's restrict to those elliptic curves of
a fixed reduction $\overline E$ over $\mathbb F_p$. We have $a_p \ne 0$, and let's assume for simplicity that the two roots
of $x^2-a_p x + p$ are in $\mathbb Z_p$, so they are $u$, $pv$ with $u$, $v \in \mathbb Z_p^\times$. Then there's a basis of
$M$ (the Fontaine-Laffaille object associated to $E[p^\infty]$) such that $\phi = \begin{bmatrix}u & \\\ & pv\end{bmatrix}$. The
possible choices of $M^1$ are precisely the lines generated by the vector $\begin{bmatrix} px\\\ 1\end{bmatrix}$, where $x
\in \mathbb Z_p$. The isomorphism class of $M$ (and thus of the Galois representation $T_p E$) only depends on the valuation
of $x$, because a diagonal change of basis leaves $\phi$ invariant. But there are uncountably many $x$ of any fixed positive
valuation (and thus deformations of $\overline E$ to $E$). By the above argument, these cannot all be isogenous.
