Edit. The short answer is that this fails for every toric variety except when $T$ equals all of $X$. I edited the original answer to prove this, and also to prove the positive result below by @S. carmeli.
Denote by $X^{(1)}$ the finite set of irreducible components of the closed subset $X\setminus T$ of $X$.
Definition 1. A principal $T$-invariant Weil divisor is an element in the kernel $\text{PDiv}^T(X)$ of the following map to the Weil divisor class group of $X$,
$$
0 \to \text{PDiv}^T(X) \to \bigoplus_{D\in X^{(1)}}\mathbb{Z}\cdot [D] \to \text{Cl}(X).
$$
Every $T$-invariant Weil divisor of $\text{PDiv}^T(X)$ is a principal Cartier divisor whose associated invertible sheaf is isomorphic to $\mathcal{O}_X$ as an $\mathcal{O}_X$-module. Moreover, since every ideal sheaf $\mathcal{O}_X(-\underline{D})\subset \mathcal{O}_X\subset i_*\mathcal{O}_T$ has a natural $T$-linearization, also the invertible sheaf of every element of $\text{PDiv}^T(X)$ has a natural $T$-linearization.
Definition 2. For every principal $T$-invariant Weil divisor, the character of each $T$-equivariant generator of the associated invertible sheaf is independent of the choice of generator up to $T$-equivariant units on $X$. The character of the principal $T$-invariant Weil divisor is the image under the associated group homomorphism,
$$
\text{char}_X: \text{PDiv}^T(X) \to \text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1))/\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1)),
$$
where we make all of the schemes pointed by distinguishing the group identity in $T\subset X$.
Lemma 3. The group homomorphism $\text{char}_X$ is an isomorphism.
Proof. Every character $$\chi:(T,1)\to (\mathbb{G}_{m,k},1),$$ is a regular morphism from $T$ to $\mathbb{A}^1_k$, and thus it determines (and is uniquely determined by) a rational function on $X$. The associated principal divisor of $\chi$ is an element of $\text{PDiv}^T(X)$ whose image under $\text{char}_X$ equals the image of $\chi$. Thus $\text{char}_X$ is surjective.
For a principal $T$-invariant Weil divisor whose image under $\text{char}_X$ is trivial, there exists a $T$-equivariant generator of the invertible sheaf whose character equals the character of a unit on $X$. Scaling the generator by the inverse of the unit, there exists a $T$-equivariant generator whose character is trivial. Thus, the principal $T$-equivariant divisor is the divisor of a $T$-equivariant rational function whose restriction to $T$ is the constant rational function $1$. Since $T$ is schematically dense in $X$, the rational function is everywhere equal to $1$. The principal divisor of the rational function $1$ is the zero divisor. QED
Proposition 4. For every pair $(\mathcal{F}_1,\mathcal{F}_2)$ of nonzero, $T$-invariant, coherent subsheaves of $i_*\mathcal{O}_T$ that are isomorphic as $\mathcal{O}_X$-modules, there exists a unique divisor $D\in \text{PDiv}^T(X)$ such that $\mathcal{F}_1$ equals $\mathcal{F}_2(\underline{D})$ as subsheaves of $i_*\mathcal{O}_T$. For every character $\chi$ of $T$ that maps to $\text{char}(D)$, the $T$-linearized $\mathcal{O}_X$-modules $\mathcal{F}_1$ and $\mathcal{F}_2\otimes \chi$ are isomorphic.
Proof. Via adjointness of $i_*$ and $i^*$, there is a natural equivalence between the set of nonzero, equivariant, coherent subsheaves of $i_*\mathcal{O}_T$ and the set of equivalence classes of pairs $(\mathcal{F},\phi)$ of a torsion-free, coherent $\mathcal{O}_X$-module $\mathcal{F}$ together with an isomorphism $$\phi:i^*\mathcal{F}\xrightarrow{\cong} \mathcal{O}_T$$ that admits a $T$-linearization. Since $X$ is normal, every such pair is of the form $(\mathcal{I}\cdot \mathcal{G},\psi|_{\mathcal{I}\cdot \mathcal{G}})$ where $(\mathcal{G},\psi)$ is such a pair with $\mathcal{G}$ reflexive, and where $\mathcal{I}$ is the ideal sheaf of a $T$-invariant closed subscheme $Z$ of $X$ that everywhere has codimension $\geq 2$ (the subscheme $Z$ is empty if $\mathcal{F}$ is already reflexive). Every $\mathcal{O}_X$-module isomorphism from $\mathcal{F}_1$ to $\mathcal{F}_2$ induces an isomorphism between the reflexive hulls $\mathcal{G}_1$ and $\mathcal{G}_2$ as well as between the ideal sheaves $\mathcal{I}_1$ and $\mathcal{I}_2$, so that also $Z_1$ equals $Z_2$ as closed subschemes of $X$. Therefore, $\mathcal{F}_1$ is isomorphic to $\mathcal{F}_2$ if and only if $\mathcal{G}_1$ is isomorphic to $\mathcal{G}_2$ and $Z_1$ equals $Z_2$. Thus, without loss of generality, assume now that $\mathcal{F}_1$ and $\mathcal{F}_2$ are reflexive.
For reflexive pairs $(\mathcal{F}_1,\phi_1)$ and $(\mathcal{F}_2,\phi_2)$, the composition $\phi_{1,2}:=\phi_1^{-1}\circ \phi_2$ is an equivariant isomorphism from $i^*\mathcal{F}_2$ to $i^*\mathcal{F}_1$. For every prime Weil divisor $D$ of $X$ contained in $X\setminus T$, both $\mathcal{F}_1$ and $\mathcal{F}_2$ are locally free of rank $1$ on a neighborhood of the generic point of $D$. Thus, the valuation of $\phi_{1,2}$ at $D$ is well-defined. Summing over all such prime Weil divisors gives a Weil divisor associated to $\phi_{1,2}$, $$\text{Div}(\phi_{1,2}) = \sum_{D\subset X\setminus T} \text{val}_D(\phi_{1,2}) \underline{D}.$$ The pushforward from the smooth locus of $X$ of the invertible sheaf of this Cartier divisor is a reflexive, coherent $\mathcal{O}_X$-module of rank $1$, say $\mathcal{O}_X(\text{Div}(\phi_{1,2}))$. Moreover, this $\mathcal{O}_X$-module has a natural $T$-linearization as a $T$-equivariant subsheaf of $i_*\mathcal{O}_T$.
Since they are reflexive, the $\mathcal{O}_X$-modules $\mathcal{F}_1$ and $\mathcal{F}_2$ are isomorphic if and only if they are isomorphic away from codimension $2$, and this holds if and only if $\mathcal{O}_X(\text{Div}(\phi_{1,2}))$ is isomorphic to $\mathcal{O}_X$, i.e., if and only if $\text{Div}(\phi_{1,2})$ is an element of $K$. In this case, $\mathcal{F}_1$ is isomorphic to $\mathcal{F}_2(\text{Div}(\phi_{1,2}))$ as subsheaves of $i_*\mathcal{O}_T$, and thus also as $T$-linearized $\mathcal{O}_X$-modules.
By the definition of $\text{char}(\text{Div}(\phi_{1,2}))$, for every character $\chi$ of $T$ mapping to $\text{char}(\text{Div}(\phi_{1,2})$, the $T$-linearized invertible sheaf $\mathcal{O}_X(\text{Div}(\phi_{1,2}))$ is isomorphic to the $T$-linearized invertible sheaf $\mathcal{O}_X\otimes_k \chi$. Thus, $\mathcal{F}_1$ is isomorphic to $\mathcal{F}_2\otimes \chi$ as a $T$-linearized, $\mathcal{O}_X$-module. QED
Corollary 5. For a toric variety $(X,T)$, the subgroup $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$ equals all of $\text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1))$ if and only if $T$ equals all of $X$. Thus, for every toric variety such that $T$ is a proper open subset of $X$, there exists a pair $(\mathcal{F}_1,\mathcal{F}_2)$ of nonzero, $T$-invariant, coherent subsheaves of $i_*\mathcal{O}_T$ that are isomorphic as $\mathcal{O}_X$-modules yet are not isomorphic as $T$-linearized, $\mathcal{O}_X$-modules.
Proof. Since $X$ is normal and $T$ is an open affine, the complement of $T$ in $X$ has pure codimension $1$. Thus, if $T$ is a proper open subset of $X$, then there exists a prime Weil divisor $D$ that is an irreducible component of $X\setminus T$. Since $X$ is normal, the local ring of $X$ at the generic point of $D$ is a discrete valuation ring; denote the associated valuation by $$\text{val}_D:k(T)^\times/k^\times\twoheadrightarrow \mathbb{Z}.$$ Since $D$ is $T$-invariant, the valuation is uniquely determined by its values on the monomials of $k[T]$. If every monomial has zero valuation, then the valuation is zero. Thus, some monomial has nonzero valuation. Up to replacing the monomial by its inverse, assume that the monomial has negative valuation. Then the monomial is an element of $\text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1))$ that is not in $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$. QED
For any particular toric variety, it is straightforward to compute $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$, $\text{PDiv}^T(X)$, and $\text{char}_X$, thus classifying all counterexamples. For instance, for $X=\mathbb{A}^n_k$, resp. $\mathbb{P}^n_k$, with its natural structure of toric variety, the group $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$ is trivial, so that the group homomorphism $\text{char}$ is an isomorphism to the character group of $T$,
$$\text{char}_{\mathbb{A}^n_k}:\text{PDiv}^T(\mathbb{A}^n_k)\xrightarrow{\cong} \text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1)),$$
$$\text{char}_{\mathbb{P}^n_k}:\text{PDiv}^T(\mathbb{P}^n_k)\xrightarrow{\cong} \text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1)).$$
Here are the simplest counterexamples. Let $X$ equal $\mathbb{A}^1_k = \text{Spec}\ k[s]$, and let $T$ equal $D(s)$ with its natural inclusion $i$, and let the group operation on $T$ be the standard one, $$s_1\bullet s_2 = s_1s_2.$$
The sheaf homomorphism $i^\#$ makes $\mathcal{O}_X$ into a $T$-equivariant sheaf. On the level of rings, the injective $k[s]$-module homomorphism is $$k[s]\hookrightarrow k[s,s^{-1}].$$ For every integer $d\geq 0$, consider the ideal $$I_d = \langle f_d \rangle \subset k[s], \ \ f_d = t^d.$$ As a $k[s]$-module, this is principal, and the set of principal generators is just the set of multiples of $f_d$ by elements of $k^\times$. In particular, this module is isomorphic to the ring as a module, so that all $\mathcal{O}_X$-modules $\widetilde{I}_d$ are isomorphic.
Since $f_d$ is a monomial, the ideal sheaf $\widetilde{I}_d \subset \mathcal{O}_X \subset i_*\mathcal{O}_T$ is equivariant. Since $\widetilde{I}_d$ is principal, it is isomorphic to $\mathcal{O}_X$ as an $\mathcal{O}_X$-module. Under every $k[s]$-module isomorphism, the generator $f_d$ of $I_d$ maps to a generator of $k[s]$, i.e., to a scalar multiple of the generator $f_0=1$. Scalar multiples of $1$ are (nonzero) $T$-invariant elements of $k[s]$. Thus, $I_d$ is equivariantly isomorphic to $k[s]$ if and only if the generator $f_d$ is a $T$-invariant element.
The coordinate $s$ on $\mathbb{A}^1_k$ identifies the $k$-algebra $\mathbb{A}^1_k(\text{Spec}\ k)$ with $k$, and that identification extends to an identification of the group $T(\text{Spec}\ k)$ with the multiplicative group $k^\times$. For every $a\in k^\times$, $$(a\cdot f_d)(s) = f_d(as) = a^d s^d = a^d f(s).$$ Thus, the character of $f_d$ is $d$ times the tautological character of $T$.
Therefore, if $d$ is nonzero, the generator $f_d$ is not $T$-invariant. Even though the equivariant subsheaf $\widetilde{I}_d\subset \mathcal{O}_X \subset i_*\mathcal{O}_T$ is isomorphic to $\mathcal{O}_X$ as a $\mathcal{O}_X$-module, there is no equivariant $\mathcal{O}_X$-module isomorphism of $\widetilde{I}_d$ with $\mathcal{O}_X$.
