Hmmm. I am not so sure that the answer is "No". In fact I would rather bet on "Yes".
Of course, I totally agree that the characteristic ideal, i.e. the ideal generated by the $p$-adic $L$-function in $\Lambda$, is not enough to determine the elliptic curve. In particular there are plenty of curves for which the $p$-adic $L$-function is a unit in $\Lambda^{\times}$.
The $p$-adic $L$-function can be viewed as a measure $\mu$ on the Galois group of $F_\infty/\mathbb{Q}$. It is build up from modular symbols of the form $\bigl[\frac{a}{p^k}\bigr]$ as $a$ and $k$ varies over all positive integers. Knowing the measure $\mu$ it is easy to extract the unit root $\alpha$ of the Frobenius at $p$ and hence the value of $a_p$. Then it is not difficult to see from the definition of $\mu$ that one can compute all the modular symbols $\bigl[\frac{a}{p^k}\bigr]$. It is true that these values do not seem to carry the value of $a_{\ell}$ for primes $\ell\neq p$ with them needed to reconstruct the complex $L$-function; we would need modular symbols with $\ell$ in the denominator and I can not see immediately how to get them from $\mu$.
Nevertheless, there are plenty of values of $a$ and $k$. And it would be a big surprise to me if there very by chance two elliptic curves such that all the values of the modular symbols $\bigl[\frac{a}{p^k}\bigr]$ would be equal. But I have no clue of how to prove this intuition.
So I ran through some examples. Let $F_{\infty}$ be the cyclotomic $\mathbb{Z}_p$-extension of $\mathbb{Q}$.
I picked a few elliptic curves of small conductor such that
$E$ has good ordinary reduction at $p$.
There are no torsion points in $E(F_{\infty})$, simply by making sure that the $\ell$-adic Galois representation is surjective for all $\ell$.
The Tamagawa numbers are all 1 for $E/F_\infty$, by imposing that the Kodaira type at all bad places is $I_1$.
The curve is not anomalous at $p$, i.e. $a_p \neq 1$. Actually, I fix $a_p$.
The Tate-Shafarevich group of $E/\mathbb{Q}$ is trivial.
The rank of $E(F_{\infty})$ is $0$. This will follow from the previous conditions if the rank of $E(\mathbb{Q})$ is $0$, since the $p$-adic $L$-function will be a unit.
Then I computed the $p$-adic $L$-functions $L_p(T)$ for these with $T$ corresponding to $1+p$ under the cyclotomic character Gal$(F_{\infty}/\mathbb{Q})$. By what I have imposed the leading term will be equal. For each $p^n$-th root of unity $\zeta$, the value of $L_p(\zeta-1)$ is, up to a power of $\alpha$ which is the same for all my curves because I fixed $a_p$, equal to the order of the Tate-Shafarevich group at the $n$-th level; at least if one believes the ($p$-adic version of the) Birch and Swinnerton-Dyer conjecture. From what I imposed, it is clear that the $p$-primary part will be trivial, but there may be different primes appearing in Sha for various curves. So there is no reason to believe that it would be easy to find two curves that have the same $p$-adic $L$-function in these family that I have chosen.
Here are some examples with $p=5$ and $a_5=-1$.
139a1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (1 + 4 \cdot 5 + O(5^2)) \cdot T + (3 + 5 + O(5^2)) \cdot T^2 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^3 + (1 + 5 + O(5^2)) \cdot T^4 + O(T^5)$
141e1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (4 + 3 \cdot 5 + O(5^2)) \cdot T + (3 \cdot 5 + O(5^2)) \cdot T^2 + (5 + O(5^2)) \cdot T^3 + (2 + 4 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5) $
346a1 $ 4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (2 + 5 + O(5^2)) \cdot T + (4 \cdot 5 + O(5^2)) \cdot T^2 + O(5^2) \cdot T^3 + (1 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$
906i1 $4 + 4 \cdot 5 + 4 \cdot 5^2 + O(5^5) + (3 + O(5^2)) \cdot T + (2 + 5 + O(5^2)) \cdot T^2 + (3 + 5 + O(5^2)) \cdot T^3 + (3 + 2 \cdot 5 + O(5^2)) \cdot T^4 + O(T^5)$
Finally a word why I believe the answer should be Yes. A big dream in the direction of BSD is to hope that there is a link between the $p$-adic and the complex $L$-function. Of course, we should believe that the order of vanishing at $s=1$ should be equal for instance (and this is only known when it is at most a simple zero). I would hope that such a link will be bijective. But that is a mere intuition and hence possibly wrong. But it also explains that the answer to this question might well be very difficult.
If I understand your question properly, then I think much is known. Let me sum up what I understand about this picture.
First a short answer to your question. Contrary to what you ask for, it is not expected that the dimension of a subspace of $H^{1}$ cut by local conditions should express the order of vanishing of the $p$-adic $L$-function.
Let us start with Bloch-Kato conjecture. This conjecture can be interpreted as a description of cohomological invariants of motives using special values of the $L$-function (many people think of it in the converse way, as description of special values of the $L$-function in terms of Galois invariants). The first question to ask is "which cohomological invariants are we trying to describe?" and the most reasonable answer is "the complex $C$ of motivic cohomology with compact support" (not known to exist in general). Then the order of vanishing of the $L$-function gives the Euler characteristic of $C\otimes_{\mathbb Q}\mathbb R$ whereas the $p$-adic valuation of the principal term of the $L$-function (divided by the period defined in Bloch-Kato) is a $\mathbb Z_{p}$-basis of the determinant of $C\otimes_{\mathbb Q}\mathbb Q_{p}$ (more precisely, of the inverse of the determinant). Even though you knew all this already, I found it necessary to recall it in order to state what forms the IMC takes in this context.
Assume now that our $p$-adic Galois representation $V$ comes from a pure motive and is crystalline at $p$ (I realize that you don't want to make such a strong assumption, but I think all I will say will continue to hold, at least conjecturally). As pointed out in comments already, and as you know, the IMC will say something about the interpolation of the Bloch-Kato conjecture in a $\mathbb Z_{p}$-extension (or more generally in a universal deformation space). I will discuss here only the case of the cyclotomic $\mathbb Z_{p}$-extension. Inside $D_{cris}(V)$ sits $D^{\phi=p^{-1}}$. Let $e$ denotes the dimension of this space over $\mathbb Q_{p}$. Then the cohomological object described by the special values of the (putative) $p$-adic $L$-function is the Selmer complex $S$ of $V$ with the unramified conditions at places $\ell≠p$ of ramifications of $V$ and with the Bloch-Kato condition at the level of complex at $p$.
Based on Bloch-Kato, we should thus expect the Euler characteristic of $S$ evaluated at a character (this is to say of $S\otimes_{\Lambda}\mathbb Z_{p}[\chi]$) to be the order of vanishing of the $p$-adic $L$-function and the $p$-adic $L$-function to give a basis of $\det_{\Lambda} S$. Alas, things are not so easy, because of the infamous trivial zeroes phenomena. So what you can show (possibly assuming plausible conjectures or restricting yourself to rank at most 2 along the way, I'll make an effort to state something really precise if you need to) is that, under Bloch-Kato, the Euler characteristic of $S\otimes_{\Lambda}\mathbb Z_{p}[\chi]$ is equal to the order of vanishing of the usual $L$-function twisted by $\chi$ (as expected) plus $e$ (this is the contribution of the trivial zeroes) $\textit{provided}$ the $\mathcal L$-invariant does not vanish (this is, or should be, equivalent to the semi-simplicity of the complex giving the local condition at $p$).
All this having been said, perhaps you want a concrete answer for a concrete representation. In that case, nothing is simpler than a brave old ordinary representation. For ordinary representation, the local condition at $p$ for the Selmer complex $S$ is simply $R\Gamma(G_{\mathbb Q_{p}},V)$. Hence, the order of vanishing of the $p$-adic $L$-function at a given $\chi$ should simply be the order of vanishing of the $L$-function plus the dimension of $H^{0}(G_{\mathbb Q_{p}},V^{*}(1)/F^{+}V^{*}(1))$ plus or minus simple terms (like the zeroes or poles of the Gamma factors). This reflects the fact that in the generic case, the order of vanishing of the $p$-adic $L$-function should be the dimension of the first cohomology of $S$ (which is not a subspace of $H^{1}$, hence my word of warning at the beginning).
Hope this helped somehow.
Now, let us move on to your second question. I think that if you knew only the IMC, then you couldn't say much about the order of vanishing part of Bloch-Kato. However, if you knew the IMC as well as non-degeneracy of the $p$-adic height pairing (required to formulate the Equivariant Tamagawa Number Conjecture) as well as the Equivariant Tamagawa Number Conjecture for each layer of the cyclotomic extension and/or the vanishing of the $\mu$-invariant, then the order part of Bloch-Kato would follow. Here is how I would try to prove this. First, I would define $S$ (no problem here,as we are in the ordinary case). Then I would construct a canonical trivialization of this complex at each finite layer using the non-degeneracy of the height pairing. Then I would use the ETNC (or I would deduce the ETNC from the IMC using the vanishing of the $\mu$-invariant) to show that the image of the determinant of $S$ at a finite layer under my canonical trivialization is really the value of the principal term of the analytic $L$-function (perhaps times the $\mathcal L$-invariant, but I would know this to be non-zero by semi-simplicity of my complexes). In this way, I would manufacture a complex $L$-function which would agree with the ordinary $L$-function at many (not necessarily classical) points (this would presumably require the IMC and ETNC not only for the cyclotomic extension but for the Hida family containing $E$) and would thus be equal to it. Now, I would know the order of vanishing of my algebraic complex $L$-function at a classical point, so I would know the order of vanishing of the complex $L$-function as well so (finally!) I could check Bloch-Kato.
So, yeah, if you knew the ETNC for the full Hida family and/or the vanishing of the $\mu$-invariant plus the non-degeneracy of the $p$-height pairing, you can, I think, collect the order part of Bloch-Kato as a bonus. Perhaps a moment of sober reflexion is in order now.
Again, hope this helped (but doubt it somehow).
Best Answer
It is false for the valuation ring in any nontrivial finite extension of $\mathbb{Q}_p$. The coefficients of the Mahler expansion of a continuous function $\mathcal{O} \to \mathbb{C}_p$ are determined by its restriction to $\mathbb{Z}_p$ (they are given as $n$-th differences of the sequence of values on nonnegative integers, in fact). But there are different continuous functions $\mathcal{O} \to \mathbb{C}_p$ with the same restriction to $\mathbb{Z}_p$.
Even worse, the Mahler expansions need not even converge because if $x$ is not in $\mathbb{Z}_p$, the binomial coefficient values may have negative valuation.
EDIT: As Kevin Buzzard and dke suggest, one can give a positive answer if your question is interpreted differently. The point of this edit is to make a few explicit remarks in these two directions.
1) If it is known in advance that $f \colon \mathcal{O} \to \mathbb{C}_p$ is represented by a single convergent power series, then the Mahler expansion of $f|_{\mathbb{Z}_p}$ converges to $f$ on all of $\mathcal{O}$. This can be deduced from the theorem that a continuous function $\mathbb{Z}_p \to \mathbb{C}_p$ is analytic if and only if the Mahler expansion coefficients $a_n$ satisfy $a_n/n! \to 0$ (see Theorem 54.4 in Ultrametric calculus: an introduction to $p$-adic analysis by W. H. Schikhof).
2) If one chooses a $\mathbb{Z}_p$-basis of $\mathcal{O}$, then $f$ can be interpreted as a continuous function $\mathbb{Z}_p^r \to \mathbb{C}_p$, and any such function has a multivariable Mahler expansion $$\sum a_n \binom{x_1}{n_1} \cdots \binom{x_r}{n_r},$$ where the sum is over tuples $n=(n_1,\ldots,n_r)$ with $n_i \in \mathbb{Z}_{\ge 0}$, and $a_n \to 0$ $p$-adically.